What is the maximum value for an int32

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I can never remember the number  I need a memory rule

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Largest negative  32bit  value   -2147483648  1  lt  lt  31   Largest positive  32bit  value   2147483647   1  lt  lt  31   Mnemonic   drunk AKA horny   drunk           Drinking age is 21 AK              AK 47 A               4  A and 4 look the same  horny           internet rule 34  if it exists  there s 18  material of it    21 47 4 years  3 years  4 years  21 47 48       36       48

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Just take any decent calculator and type in  7FFFFFFF  in hex mode  then switch to decimal   2147483647

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It s 10 digits  so pretend it s a phone number  assuming you re in the US   214-748-3647  I don t recommend calling it

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if you can remember the entire Pi number  then the number you are looking for is at the position 1 867 996 680 till 1 867 996 689 of the decimal digits of Pi      The numeric string 2147483647 appears at the 1 867 996 680 decimal digit of Pi  3 14      86181221809936452346214748364710527835665425671614      source  http   www subidiom com pi

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2  x y    2 x   2 y  2 10   1 000 2 20   1 000 000 2 30   1 000 000 000 2 40   1 000 000 000 000  etc    2 1   2 2 2   4 2 3   8 2 4   16 2 5   32 2 6   64 2 7   128 2 8   256 2 9   512   So  2 31  signed int max  is 2 30  about 1 billion  times 2 1  2   or about 2 billion  And 2 32 is 2 30   2 2 or about 4 billion  This method of approximation is accurate enough even out to around 2 64  where the error grows to about 15     If you need an exact answer then you should pull up a calculator   Handy word-aligned capacity approximations    2 16    64 thousand    uint16 2 32    4 billion    uint32  IPv4  unixtime 2 64    16 quintillion  aka 16 billion billions or 16 million trillions     uint64   bigint  2 128    256 quintillion quintillion  aka 256 trillion trillion trillions     IPv6  GUID

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It s 231  minus  1  32 bits  one is used for sign    If you want an approximate value  use 210   1024  asymp  103  so 231  asymp  2 109  If you want to compute an exact value by hand  use exponentiation by squaring to get to 232   2 25  and divide by two  You only need to square five times to get 232   2 2   4 4 4   16 16 16   256 256 256   25 25 100   2 250 6   36   62500   3000   36   65536 65536 65536  65000 65000   2 65000 536   536 536     4225000000   130000 536    250000   3600   36 36    4225000000   69680000   250000   3600   1296   4294967296   dividing this by two and subtracting one gives you 2 147 483 647  If you don t need all digits  but only want say  the first three significant digits  the computations on each squaring step are very easy

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this is how i do it to remember 2 147 483 647  To a far savannah quarter optimus trio hexed forty septenary  2 - To 1 - A 4 - Far 7 - Savannah 4 - Quarter 8 - Optimus 3 - Trio 6 - Hexed 4 - Forty 7 - Septenary

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2  x y    2 x   2 y  2 10   1 000 2 20   1 000 000 2 30   1 000 000 000 2 40   1 000 000 000 000  etc    2 1   2 2 2   4 2 3   8 2 4   16 2 5   32 2 6   64 2 7   128 2 8   256 2 9   512   So  2 31  signed int max  is 2 30  about 1 billion  times 2 1  2   or about 2 billion  And 2 32 is 2 30   2 2 or about 4 billion  This method of approximation is accurate enough even out to around 2 64  where the error grows to about 15     If you need an exact answer then you should pull up a calculator   Handy word-aligned capacity approximations    2 16    64 thousand    uint16 2 32    4 billion    uint32  IPv4  unixtime 2 64    16 quintillion  aka 16 billion billions or 16 million trillions     uint64   bigint  2 128    256 quintillion quintillion  aka 256 trillion trillion trillions     IPv6  GUID

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That s how I remembered 2147483647    214 - because 2 14 is approximately pi-1 48   6 8 64   8 8   Write these horizontally   214 48 64  and insert                7  3  7 - which is Boeing s airliner jet  thanks  sgorozco    Now you ve got 2147483647   Hope this helps at least a bit

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Using Java 9 s REPL  jshell     jshell    Welcome to JShell -- Version 9-Debian  jshell gt  System out println Integer MAX VALUE  2147483647

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Just take any decent calculator and type in  7FFFFFFF  in hex mode  then switch to decimal   2147483647

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If a huge integer isn t recalled  you recall this mnemonic    Now count the letters in each word

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That s how I remembered 2147483647    214 - because 2 14 is approximately pi-1 48   6 8 64   8 8   Write these horizontally   214 48 64  and insert                7  3  7 - which is Boeing s airliner jet  thanks  sgorozco    Now you ve got 2147483647   Hope this helps at least a bit

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With Groovy on the path   groovy -e   println Integer MAX VALUE      Groovy is extremely useful for quick reference  within a Java context

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It is very easy to remember  In hexadecimal one digit is 4 bits  So for unsigned int write 0x and 8 fs  0xffffffff  into a Python or Ruby shell to get the value in base 10  If you need the signed value  just remember that the highest bit is used as the sign  So you have to leave that out  You only need to remember that the number where the lower 3 bits are 1 and the 4th bit is 0 equals 7  so write 0x7fffffff into a Python or Ruby shell  You could also write 0x100000000 - 1 and 0x80000000 - 1  if that is more easy to you to remember

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Assuming  NET -  Console WriteLine Int32 MaxValue

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Rather than think of it as one big number  try breaking it down and looking for associated ideas eg    2 maximum snooker breaks  a maximum break is 147  4 years  48 months  3 years  36 months  4 years  48 months    The above applies to the biggest negative number  positive is that minus one   Maybe the above breakdown will be no more memorable for you  it s hardly exciting is it    but hopefully you can come up with some ideas that are

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In general you could do a simple operation which reflects the very nature of a Int32  fill all the available bits with 1 s - That is something which you can hold easily in your memory  It works basically the same way in most languages  but i m going with Python for the example   max   0 bits    1    31   Generate a  bit array  filled with 1 s for bit in bits      max    max  lt  lt  1    bit   max is now 2147483647   For unsigned Int32 s  make it 32 instead of 31 1 s   But since there are posted a few more adventurous approaches  i began to think of formulas  just for the fun of it     Formula 1  Numbers are concatenated if no operator is given    a   4 b   8 ba a ab-1 ab ab-a-b ab-1   Python quickcheck  a   4 b   8 ab   int   d d     a  b   ba   int   d d     b  a     d d d d d     ba a  ab-1  ab  ab-a-b  ab-1    gives  2147483647    Formula 2   x   48 x 2-3 x-1 x x 3 4 x-1   Python quickcheck  x   48   d d d d d     x 2-3  x-1  x  x 3 4  x-1     gives  2147483647

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Just remember that 2  10 x  is approximately 10  3 x  - you re probably already used to this with kilobytes kibibytes etc   That is   2 10   1024                   one thousand 2 20   1024 2   1048576       one million 2 30   1024 3   1073741824    one billion   Since an int uses 31 bits     1 bit for the sign   just double 2 30 to get approximately 2 billion   For an unsigned int using 32 bits  double again for 4 billion   The error factor gets higher the larger you go of course  but you don t need the exact value memorised  If you need it  you should be using a pre-defined constant for it anyway    The approximate value is good enough for noticing when something might be a dangerously close to overflowing

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Interestingly  Int32 MaxValue has more characters than 2 147 486 647   But then again  we do have code completion   So I guess all we really have to memorize is Int3 lt period gt M lt enter gt   which is only 6 characters to type in visual studio   UPDATE For some reason I was downvoted  The only reason I can think of is that they didn t understand my first statement    Int32 MaxValue  takes at most 14 characters to type  2 147 486 647 takes either 10 or 13 characters to type depending on if you put the commas in or not

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Here s a mnemonic for remembering 2  31  subtract one to get the maximum integer value   a 1 b 2 c 3 d 4 e 5 f 6 g 7 h 8 i 9  Boys And Dogs Go Duck Hunting  Come Friday Ducks Hide 2    1   4    7  4    8        3    6      4     8   I ve used the powers of two up to 18 often enough to remember them  but even I haven t bothered memorizing 2  31  It s too easy to calculate as needed or use a constant  or estimate as 2G

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With Groovy on the path   groovy -e   println Integer MAX VALUE      Groovy is extremely useful for quick reference  within a Java context

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The most correct answer I can think of is Int32 MaxValue

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You will find in binary the maximum value of an Int32 is 1111111111111111111111111111111 but in ten based you will find it is 2147483647 or 2 31-1 or Int32 MaxValue

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The most correct answer I can think of is Int32 MaxValue

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Just remember that 2  10 x  is approximately 10  3 x  - you re probably already used to this with kilobytes kibibytes etc   That is   2 10   1024                   one thousand 2 20   1024 2   1048576       one million 2 30   1024 3   1073741824    one billion   Since an int uses 31 bits     1 bit for the sign   just double 2 30 to get approximately 2 billion   For an unsigned int using 32 bits  double again for 4 billion   The error factor gets higher the larger you go of course  but you don t need the exact value memorised  If you need it  you should be using a pre-defined constant for it anyway    The approximate value is good enough for noticing when something might be a dangerously close to overflowing

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Rather than think of it as one big number  try breaking it down and looking for associated ideas eg    2 maximum snooker breaks  a maximum break is 147  4 years  48 months  3 years  36 months  4 years  48 months    The above applies to the biggest negative number  positive is that minus one   Maybe the above breakdown will be no more memorable for you  it s hardly exciting is it    but hopefully you can come up with some ideas that are

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It is very easy to remember  In hexadecimal one digit is 4 bits  So for unsigned int write 0x and 8 fs  0xffffffff  into a Python or Ruby shell to get the value in base 10  If you need the signed value  just remember that the highest bit is used as the sign  So you have to leave that out  You only need to remember that the number where the lower 3 bits are 1 and the 4th bit is 0 equals 7  so write 0x7fffffff into a Python or Ruby shell  You could also write 0x100000000 - 1 and 0x80000000 - 1  if that is more easy to you to remember

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If you happen to know your ASCII table off by heart and not MaxInt    GH6G   21 47 48 36 47

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Assuming  NET -  Console WriteLine Int32 MaxValue

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Largest negative  32bit  value   -2147483648  1  lt  lt  31   Largest positive  32bit  value   2147483647   1  lt  lt  31   Mnemonic   drunk AKA horny   drunk           Drinking age is 21 AK              AK 47 A               4  A and 4 look the same  horny           internet rule 34  if it exists  there s 18  material of it    21 47 4 years  3 years  4 years  21 47 48       36       48

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2147483647  Here s what you need to remember    It s 2 billion  The next three triplets are increasing like so  100s  400s  600s The first and the last triplet need 3 added to them so they get rounded up to 50  eg 147   3   150  amp  647   3   650  The second triplet needs 3 subtracted from it to round it down to 80  eg 483 - 3   480    Hence 2  147  483  647

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It s 10 digits  so pretend it s a phone number  assuming you re in the US   214-748-3647  I don t recommend calling it

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Int32 means you have 32 bits available to store your number  The highest bit is the sign-bit  this indicates if the number is positive or negative  So you have 2 31 bits for positive and negative numbers    With zero being a positive number you get the logical range of  mentioned before    2147483647 to -2147483648  If you think that is to small  use Int64    9223372036854775807 to -9223372036854775808  And why the hell you want to remember this number  To use in your code  You should always use Int32 MaxValue or Int32 MinValue in your code since these are static values  within the  net core  and thus faster in use than creating a new int with code   My statement  if know this number by memory   you re just showing off

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In general you could do a simple operation which reflects the very nature of a Int32  fill all the available bits with 1 s - That is something which you can hold easily in your memory  It works basically the same way in most languages  but i m going with Python for the example   max   0 bits    1    31   Generate a  bit array  filled with 1 s for bit in bits      max    max  lt  lt  1    bit   max is now 2147483647   For unsigned Int32 s  make it 32 instead of 31 1 s   But since there are posted a few more adventurous approaches  i began to think of formulas  just for the fun of it     Formula 1  Numbers are concatenated if no operator is given    a   4 b   8 ba a ab-1 ab ab-a-b ab-1   Python quickcheck  a   4 b   8 ab   int   d d     a  b   ba   int   d d     b  a     d d d d d     ba a  ab-1  ab  ab-a-b  ab-1    gives  2147483647    Formula 2   x   48 x 2-3 x-1 x x 3 4 x-1   Python quickcheck  x   48   d d d d d     x 2-3  x-1  x  x 3 4  x-1     gives  2147483647

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Rather than think of it as one big number  try breaking it down and looking for associated ideas eg    2 maximum snooker breaks  a maximum break is 147  4 years  48 months  3 years  36 months  4 years  48 months    The above applies to the biggest negative number  positive is that minus one   Maybe the above breakdown will be no more memorable for you  it s hardly exciting is it    but hopefully you can come up with some ideas that are

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Remember this  21 IQ ITEM 47  It can be de-encoded with any phone pad  or you can just write one down yourself on a paper   In order to remember  21 IQ ITEM 47   I would go with  Hitman Codename 47 had 21 missions  which were each IQ ITEM s by themselves    Or  I clean teeth at 21 47 every day  because I have high IQ and don t like items in my mouth

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Try in Python    gt  gt  gt  int  1    31  base 2  2147483647

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2  x y    2 x   2 y  2 10   1 000 2 20   1 000 000 2 30   1 000 000 000 2 40   1 000 000 000 000  etc    2 1   2 2 2   4 2 3   8 2 4   16 2 5   32 2 6   64 2 7   128 2 8   256 2 9   512   So  2 31  signed int max  is 2 30  about 1 billion  times 2 1  2   or about 2 billion  And 2 32 is 2 30   2 2 or about 4 billion  This method of approximation is accurate enough even out to around 2 64  where the error grows to about 15     If you need an exact answer then you should pull up a calculator   Handy word-aligned capacity approximations    2 16    64 thousand    uint16 2 32    4 billion    uint32  IPv4  unixtime 2 64    16 quintillion  aka 16 billion billions or 16 million trillions     uint64   bigint  2 128    256 quintillion quintillion  aka 256 trillion trillion trillions     IPv6  GUID

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It s 2 147 483 647  Easiest way to memorize it is via a tattoo

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If a huge integer isn t recalled  you recall this mnemonic    Now count the letters in each word

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Remember this  21 IQ ITEM 47  It can be de-encoded with any phone pad  or you can just write one down yourself on a paper   In order to remember  21 IQ ITEM 47   I would go with  Hitman Codename 47 had 21 missions  which were each IQ ITEM s by themselves    Or  I clean teeth at 21 47 every day  because I have high IQ and don t like items in my mouth

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In C use INT32 MAX after  include  lt stdint h gt   In C   use INT32 MAX after  include  lt cstdint gt    Or INT MAX for platform-specific size or UINT32 MAX or UINT MAX for unsigned int   See http   www cplusplus com reference cstdint  and http   www cplusplus com reference climits    Or sizeof int

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Just take any decent calculator and type in  7FFFFFFF  in hex mode  then switch to decimal   2147483647

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It s 10 digits  so pretend it s a phone number  assuming you re in the US   214-748-3647  I don t recommend calling it

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You will find in binary the maximum value of an Int32 is 1111111111111111111111111111111 but in ten based you will find it is 2147483647 or 2 31-1 or Int32 MaxValue

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It s about 2 1   10 9  No need to know the exact 2  31  - 1   2 147 483 647   C  You can find it in C like that    include  lt stdio h gt   include  lt limits h gt   main         printf  max int  t t i n   INT MAX       printf  max unsigned int  t u n   UINT MAX       gives  well  without the     max int           2 147 483 647 max unsigned int  4 294 967 295   C   11  std  cout  lt  lt  std  numeric limits lt int gt   max    lt  lt    n   std  cout  lt  lt  std  numeric limits lt unsigned int gt   max    lt  lt    n     Java  You can get this with Java  too   System out println Integer MAX VALUE     But keep in mind that Java integers are always signed   Python 2  Python has arbitrary precision integers  But in Python 2  they are mapped to C integers  So you can do this   import sys sys maxint  gt  gt  gt  2147483647 sys maxint   1  gt  gt  gt  2147483648L   So Python switches to long when the integer gets bigger than 2 31 -1

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If you happen to know your ASCII table off by heart and not MaxInt    GH6G   21 47 48 36 47

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The easiest way to do this for integers is to use hexadecimal  provided that there isn t something like Int maxInt    The reason is this   Max unsigned values  8-bit 0xFF 16-bit 0xFFFF 32-bit 0xFFFFFFFF 64-bit 0xFFFFFFFFFFFFFFFF 128-bit 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF   Signed values  using 7F as the max signed value  8-bit 0x7F 16-bit 0x7FFF 32-bit 0x7FFFFFFF 64-bit 0x7FFFFFFFFFFFFFFF   Signed values  using 80 as the max signed value  8-bit 0x80 16-bit 0x8000 32-bit 0x80000000 64-bit 0x8000000000000000   How does this work  This is very similar to the binary tactic  and each hex digit is exactly 4 bits  Also  a lot of compilers support hex a lot better than they support binary    F hex to binary  1111 8 hex to binary  1000 7 hex to binary  0111 0 hex to binary  0000   So 7F is equal to 01111111   7FFF is equal to 0111111111111111  Also  if you are using this for  insanely-high constant   7F    is safe hex  but it s easy enough to try out 7F and 80 and just print them to your screen to see which one it is   0x7FFF   0x0001   0x8000  so your loss is only one number  so using 0x7F    usually isn t a bad tradeoff for more reliable code  especially once you start using 32-bits or more

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if you can remember the entire Pi number  then the number you are looking for is at the position 1 867 996 680 till 1 867 996 689 of the decimal digits of Pi      The numeric string 2147483647 appears at the 1 867 996 680 decimal digit of Pi  3 14      86181221809936452346214748364710527835665425671614      source  http   www subidiom com pi

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It s 2 147 483 647  Easiest way to memorize it is via a tattoo

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What do you mean  It should be easy enough to remember that it is 2 32  If you want a rule to memorize the value of that number  a handy rule of thumb is for converting between binary and decimal in general   2 10   1000  which means 2 20   1 000 000  and 2 30   1 000 000 000  Double that  2 31  is rounghly 2 billion  and doubling that again  2 32  is 4 billion   It s an easy way to get a rough estimate of any binary number  10 zeroes in binary becomes 3 zeroes in decimal

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2147483647  Here s what you need to remember    It s 2 billion  The next three triplets are increasing like so  100s  400s  600s The first and the last triplet need 3 added to them so they get rounded up to 50  eg 147   3   150  amp  647   3   650  The second triplet needs 3 subtracted from it to round it down to 80  eg 483 - 3   480    Hence 2  147  483  647

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The most correct answer I can think of is Int32 MaxValue

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Interestingly  Int32 MaxValue has more characters than 2 147 486 647   But then again  we do have code completion   So I guess all we really have to memorize is Int3 lt period gt M lt enter gt   which is only 6 characters to type in visual studio   UPDATE For some reason I was downvoted  The only reason I can think of is that they didn t understand my first statement    Int32 MaxValue  takes at most 14 characters to type  2 147 486 647 takes either 10 or 13 characters to type depending on if you put the commas in or not

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Anyway  take this regex  it determines if the string contains a non-negative Integer in  decimal form that is also not greater than Int32 MaxValue     0-9  1 9   0-1  0-9  1 8  20 0-9  1 8  21 0-3  0-9  1 7  214 0-6  0-9  1 7  2147 0-3  0-9  1 6  21474 0-7  0-9  1 5  214748 0-2  0-9  1 4  2147483 0-5  0-9  1 3  21474836 0-3  0-9  1 2  214748364 0-7   Maybe it would help you to remember

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It s about 2 1   10 9  No need to know the exact 2  31  - 1   2 147 483 647   C  You can find it in C like that    include  lt stdio h gt   include  lt limits h gt   main         printf  max int  t t i n   INT MAX       printf  max unsigned int  t u n   UINT MAX       gives  well  without the     max int           2 147 483 647 max unsigned int  4 294 967 295   C   11  std  cout  lt  lt  std  numeric limits lt int gt   max    lt  lt    n   std  cout  lt  lt  std  numeric limits lt unsigned int gt   max    lt  lt    n     Java  You can get this with Java  too   System out println Integer MAX VALUE     But keep in mind that Java integers are always signed   Python 2  Python has arbitrary precision integers  But in Python 2  they are mapped to C integers  So you can do this   import sys sys maxint  gt  gt  gt  2147483647 sys maxint   1  gt  gt  gt  2147483648L   So Python switches to long when the integer gets bigger than 2 31 -1

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Just remember that 2  10 x  is approximately 10  3 x  - you re probably already used to this with kilobytes kibibytes etc   That is   2 10   1024                   one thousand 2 20   1024 2   1048576       one million 2 30   1024 3   1073741824    one billion   Since an int uses 31 bits     1 bit for the sign   just double 2 30 to get approximately 2 billion   For an unsigned int using 32 bits  double again for 4 billion   The error factor gets higher the larger you go of course  but you don t need the exact value memorised  If you need it  you should be using a pre-defined constant for it anyway    The approximate value is good enough for noticing when something might be a dangerously close to overflowing

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It s 10 digits  so pretend it s a phone number  assuming you re in the US   214-748-3647  I don t recommend calling it

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In Objective-C  iOS  amp  OSX   just remember these macros    define INT8 MAX         127  define INT16 MAX        32767  define INT32 MAX        2147483647  define INT64 MAX        9223372036854775807LL   define UINT8 MAX         255  define UINT16 MAX        65535  define UINT32 MAX        4294967295U  define UINT64 MAX        18446744073709551615ULL

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To never forget the maximum value of any type   If it has 32 bits  the largest possible value would be the 32 bits with the number 1     The result would be 4294967295 in decimal     But  as there is also the representation of negative numbers  divide 4294967295 by 2 and get 2147483647   Therefore  a 32-bit integer is capable of representing -2147483647 to 2147483647

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It s 231  minus  1  32 bits  one is used for sign    If you want an approximate value  use 210   1024  asymp  103  so 231  asymp  2 109  If you want to compute an exact value by hand  use exponentiation by squaring to get to 232   2 25  and divide by two  You only need to square five times to get 232   2 2   4 4 4   16 16 16   256 256 256   25 25 100   2 250 6   36   62500   3000   36   65536 65536 65536  65000 65000   2 65000 536   536 536     4225000000   130000 536    250000   3600   36 36    4225000000   69680000   250000   3600   1296   4294967296   dividing this by two and subtracting one gives you 2 147 483 647  If you don t need all digits  but only want say  the first three significant digits  the computations on each squaring step are very easy

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First write out 47 twice   you like Agent 47  right    keeping spaces as shown  each dash is a slot for a single digit  First 2 slots  then 4   --47----47   Think you have 12 in hand  because 12   a dozen   Multiply it by 4  first digit of Agent 47 s number  i e  47  and place the result to the right of first pair you already have  12   4   48 --4748--47  lt -- after placing 48 to the right of first 47   Then multiply 12 by 3  in order to make second digit of Agent 47 s number  which is 7  you need 7 - 4   3  and put the result to the right of the first 2 pairs  the last pair-slot  12   3   36 --47483647  lt -- after placing 36 to the right of first two pairs   Finally drag digits one by one from your hand starting from right-most digit  2 in this case  and place them in the first empty slot you get  2-47483647  lt -- after placing 2 2147483647  lt -- after placing 1   There you have it  For negative limit  you can think of that as 1 more in absolute value than the positive limit   Practise a few times  and you will get the hang of it

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I made a couple genius methods in C  that you can take advantage of in your production environment   public static int GetIntMaxValueGenius1         int n   0      while    n  gt  0          return --n     public static int GetIntMaxValueGenius2         int n   0      try               while  true              n   checked n   1             catch         return n

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In Objective-C  iOS  amp  OSX   just remember these macros    define INT8 MAX         127  define INT16 MAX        32767  define INT32 MAX        2147483647  define INT64 MAX        9223372036854775807LL   define UINT8 MAX         255  define UINT16 MAX        65535  define UINT32 MAX        4294967295U  define UINT64 MAX        18446744073709551615ULL

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If you think the value is too hard to remember in base 10  try base 2  1111111111111111111111111111111

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Well  it has 32 bits and hence can store 2 32 different values  Half of those are negative   The solution is 2 147 483 647  And the lowest is -2 147 483 648    Notice that there is one more negative value

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The easiest way to remember is to look at  std  numeric limits lt  int  gt   max    For example  from MSDN       numeric limits max cpp   include  lt iostream gt   include  lt limits gt   using namespace std   int main        cout  lt  lt   The maximum value for type float is              lt  lt  numeric limits lt float gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type double is              lt  lt  numeric limits lt double gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type int is              lt  lt  numeric limits lt int gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type short int is              lt  lt  numeric limits lt short int gt   max             lt  lt  endl

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Int32 means you have 32 bits available to store your number  The highest bit is the sign-bit  this indicates if the number is positive or negative  So you have 2 31 bits for positive and negative numbers    With zero being a positive number you get the logical range of  mentioned before    2147483647 to -2147483648  If you think that is to small  use Int64    9223372036854775807 to -9223372036854775808  And why the hell you want to remember this number  To use in your code  You should always use Int32 MaxValue or Int32 MinValue in your code since these are static values  within the  net core  and thus faster in use than creating a new int with code   My statement  if know this number by memory   you re just showing off

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2GB   is there a minimum length for answers

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With Groovy on the path   groovy -e   println Integer MAX VALUE      Groovy is extremely useful for quick reference  within a Java context

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I made a couple genius methods in C  that you can take advantage of in your production environment   public static int GetIntMaxValueGenius1         int n   0      while    n  gt  0          return --n     public static int GetIntMaxValueGenius2         int n   0      try               while  true              n   checked n   1             catch         return n

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max signed 32 bit num   1  lt  lt  31 - 1      alternatively   1  lt  lt  31    A compiler should optimize it anyway   I prefer 1  lt  lt  31 - 1 over   0x7fffffff because you don t need count fs  unsigned  pow  2  31     - 1 because you don t need  lt math h gt

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Well  aside from jokes  if you re really looking for a useful memory rule  there is one that I always use for remembering big numbers   You need to break down your number into parts from 3-4 digits and remember them visually using projection on your cell phone keyboard  It s easier to show on a picture     As you can see  from now on you just have to remember 3 shapes  2 of them looks like a Tetris L and one looks like a tick  Which is definitely much easier than memorizing a 10-digit number   When you need to recall the number just recall the shapes  imagine look on a phone keyboard and project the shapes on it  Perhaps initially you ll have to look at the keyboard but after just a bit of practice  you ll remember that numbers are going from top-left to bottom-right so you will be able to simply imagine it in your head   Just make sure you remember the direction of shapes and the number of digits in each shape  for instance  in 2147483647 example we have a 4-digit Tetris L and a 3-digit L    You can use this technique to easily remember any important numbers  for instance  I remembered my 16-digit credit card number etc

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Using Java 9 s REPL  jshell     jshell    Welcome to JShell -- Version 9-Debian  jshell gt  System out println Integer MAX VALUE  2147483647

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This is how I remember    In hex  a digit represents four bits  so 4   8   32  so the max signed 32 bit int is   0xFFFFFFFF  gt  gt  1     gt  2147483647

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With Groovy on the path   groovy -e   println Integer MAX VALUE      Groovy is extremely useful for quick reference  within a Java context

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Well  it has 32 bits and hence can store 2 32 different values  Half of those are negative   The solution is 2 147 483 647  And the lowest is -2 147 483 648    Notice that there is one more negative value

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32 bits  one for the sign  31 bits of information   2 31 - 1   2147483647   Why -1  Because the first is zero  so the greatest is the count minus one   EDIT for cantfindaname88  The count is 2 31 but the greatest can t be 2147483648  2 31  because we count from 0  not 1   Rank   1 2 3 4 5 6     2147483648 Number 0 1 2 3 4 5     2147483647   Another explanation with only 3 bits   1 for the sign  2 for the information  2 2 - 1   3   Below all the possible values with 3 bits   2 3   8 values   1  100    gt  -4 2  101    gt  -3 3  110    gt  -2 4  111    gt  -1 5  000    gt   0 6  001    gt   1 7  010    gt   2 8  011    gt   3

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Just remember that it s the eighth Mersenne prime   If that s too hard  it s also the third of only four known double Mersenne primes   Edit per comment request   The Euclid-Euler theorem states that every even perfect number has the form 2  n - 1   2 n - 1   where 2 n - 1 is a prime number  The prime numbers of the form 2 n - 1 are known as Mersenne primes  and require n itself to be prime   We know that the length of an INT32 is of course 32 bits  Given the generally accepted understanding of 2 s complement  a signed INT32 is 32 bits - 1 bit    To find the magnitude of a binary number with a given number of bits we generally raise 2 to the power n  minus 1  where n is equal to the number of bits   Thus the magnitude calculation is 2  32 - 1  - 1   2 31 - 1  31 is prime and as outlined above  prime numbers of this form are Mersenne primes  We can prove it is the eight of such simply by counting them  For further details  please ask Euler  or maybe Bernoulli  to whom he wrote about them    See  https   books google ie books id x7p4tCPPuXoC amp printsec frontcover amp dq 9780883853283 amp hl en amp sa X amp ved 0ahUKEwilzbORuJLdAhUOiaYKHcsZD-EQ6AEIKTAA v onepage amp q 9780883853283 amp f false

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Here s a mnemonic for remembering 2  31  subtract one to get the maximum integer value   a 1 b 2 c 3 d 4 e 5 f 6 g 7 h 8 i 9  Boys And Dogs Go Duck Hunting  Come Friday Ducks Hide 2    1   4    7  4    8        3    6      4     8   I ve used the powers of two up to 18 often enough to remember them  but even I haven t bothered memorizing 2  31  It s too easy to calculate as needed or use a constant  or estimate as 2G

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Anyway  take this regex  it determines if the string contains a non-negative Integer in  decimal form that is also not greater than Int32 MaxValue     0-9  1 9   0-1  0-9  1 8  20 0-9  1 8  21 0-3  0-9  1 7  214 0-6  0-9  1 7  2147 0-3  0-9  1 6  21474 0-7  0-9  1 5  214748 0-2  0-9  1 4  2147483 0-5  0-9  1 3  21474836 0-3  0-9  1 2  214748364 0-7   Maybe it would help you to remember

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this is how i do it to remember 2 147 483 647  To a far savannah quarter optimus trio hexed forty septenary  2 - To 1 - A 4 - Far 7 - Savannah 4 - Quarter 8 - Optimus 3 - Trio 6 - Hexed 4 - Forty 7 - Septenary

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In C use INT32 MAX after  include  lt stdint h gt   In C   use INT32 MAX after  include  lt cstdint gt    Or INT MAX for platform-specific size or UINT32 MAX or UINT MAX for unsigned int   See http   www cplusplus com reference cstdint  and http   www cplusplus com reference climits    Or sizeof int

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Try in Python    gt  gt  gt  int  1    31  base 2  2147483647

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Assuming  NET -  Console WriteLine Int32 MaxValue

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Just remember that 2  10 x  is approximately 10  3 x  - you re probably already used to this with kilobytes kibibytes etc   That is   2 10   1024                   one thousand 2 20   1024 2   1048576       one million 2 30   1024 3   1073741824    one billion   Since an int uses 31 bits     1 bit for the sign   just double 2 30 to get approximately 2 billion   For an unsigned int using 32 bits  double again for 4 billion   The error factor gets higher the larger you go of course  but you don t need the exact value memorised  If you need it  you should be using a pre-defined constant for it anyway    The approximate value is good enough for noticing when something might be a dangerously close to overflowing

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The easiest way to do this for integers is to use hexadecimal  provided that there isn t something like Int maxInt    The reason is this   Max unsigned values  8-bit 0xFF 16-bit 0xFFFF 32-bit 0xFFFFFFFF 64-bit 0xFFFFFFFFFFFFFFFF 128-bit 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF   Signed values  using 7F as the max signed value  8-bit 0x7F 16-bit 0x7FFF 32-bit 0x7FFFFFFF 64-bit 0x7FFFFFFFFFFFFFFF   Signed values  using 80 as the max signed value  8-bit 0x80 16-bit 0x8000 32-bit 0x80000000 64-bit 0x8000000000000000   How does this work  This is very similar to the binary tactic  and each hex digit is exactly 4 bits  Also  a lot of compilers support hex a lot better than they support binary    F hex to binary  1111 8 hex to binary  1000 7 hex to binary  0111 0 hex to binary  0000   So 7F is equal to 01111111   7FFF is equal to 0111111111111111  Also  if you are using this for  insanely-high constant   7F    is safe hex  but it s easy enough to try out 7F and 80 and just print them to your screen to see which one it is   0x7FFF   0x0001   0x8000  so your loss is only one number  so using 0x7F    usually isn t a bad tradeoff for more reliable code  especially once you start using 32-bits or more

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If you think the value is too hard to remember in base 10  try base 2  1111111111111111111111111111111

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It s 2 147 483 647  Easiest way to memorize it is via a tattoo

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Rather than think of it as one big number  try breaking it down and looking for associated ideas eg    2 maximum snooker breaks  a maximum break is 147  4 years  48 months  3 years  36 months  4 years  48 months    The above applies to the biggest negative number  positive is that minus one   Maybe the above breakdown will be no more memorable for you  it s hardly exciting is it    but hopefully you can come up with some ideas that are

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Well  it has 32 bits and hence can store 2 32 different values  Half of those are negative   The solution is 2 147 483 647  And the lowest is -2 147 483 648    Notice that there is one more negative value

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2  x y    2 x   2 y  2 10   1 000 2 20   1 000 000 2 30   1 000 000 000 2 40   1 000 000 000 000  etc    2 1   2 2 2   4 2 3   8 2 4   16 2 5   32 2 6   64 2 7   128 2 8   256 2 9   512   So  2 31  signed int max  is 2 30  about 1 billion  times 2 1  2   or about 2 billion  And 2 32 is 2 30   2 2 or about 4 billion  This method of approximation is accurate enough even out to around 2 64  where the error grows to about 15     If you need an exact answer then you should pull up a calculator   Handy word-aligned capacity approximations    2 16    64 thousand    uint16 2 32    4 billion    uint32  IPv4  unixtime 2 64    16 quintillion  aka 16 billion billions or 16 million trillions     uint64   bigint  2 128    256 quintillion quintillion  aka 256 trillion trillion trillions     IPv6  GUID

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Well  aside from jokes  if you re really looking for a useful memory rule  there is one that I always use for remembering big numbers   You need to break down your number into parts from 3-4 digits and remember them visually using projection on your cell phone keyboard  It s easier to show on a picture     As you can see  from now on you just have to remember 3 shapes  2 of them looks like a Tetris L and one looks like a tick  Which is definitely much easier than memorizing a 10-digit number   When you need to recall the number just recall the shapes  imagine look on a phone keyboard and project the shapes on it  Perhaps initially you ll have to look at the keyboard but after just a bit of practice  you ll remember that numbers are going from top-left to bottom-right so you will be able to simply imagine it in your head   Just make sure you remember the direction of shapes and the number of digits in each shape  for instance  in 2147483647 example we have a 4-digit Tetris L and a 3-digit L    You can use this technique to easily remember any important numbers  for instance  I remembered my 16-digit credit card number etc

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This is how I remember    In hex  a digit represents four bits  so 4   8   32  so the max signed 32 bit int is   0xFFFFFFFF  gt  gt  1     gt  2147483647

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To never forget the maximum value of any type   If it has 32 bits  the largest possible value would be the 32 bits with the number 1     The result would be 4294967295 in decimal     But  as there is also the representation of negative numbers  divide 4294967295 by 2 and get 2147483647   Therefore  a 32-bit integer is capable of representing -2147483647 to 2147483647

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Just take any decent calculator and type in  7FFFFFFF  in hex mode  then switch to decimal   2147483647

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Assuming  NET -  Console WriteLine Int32 MaxValue

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2GB   is there a minimum length for answers

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2GB   is there a minimum length for answers

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32 bits  one for the sign  31 bits of information   2 31 - 1   2147483647   Why -1  Because the first is zero  so the greatest is the count minus one   EDIT for cantfindaname88  The count is 2 31 but the greatest can t be 2147483648  2 31  because we count from 0  not 1   Rank   1 2 3 4 5 6     2147483648 Number 0 1 2 3 4 5     2147483647   Another explanation with only 3 bits   1 for the sign  2 for the information  2 2 - 1   3   Below all the possible values with 3 bits   2 3   8 values   1  100    gt  -4 2  101    gt  -3 3  110    gt  -2 4  111    gt  -1 5  000    gt   0 6  001    gt   1 7  010    gt   2 8  011    gt   3

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The best rule to memorize it is  21  magic number   47  just remember it  48  sequential   36  21   15  both magics   47 again  Also it is easier to remember 5 pairs than 10 digits

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Well  it has 32 bits and hence can store 2 32 different values  Half of those are negative   The solution is 2 147 483 647  And the lowest is -2 147 483 648    Notice that there is one more negative value

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max signed 32 bit num   1  lt  lt  31 - 1      alternatively   1  lt  lt  31    A compiler should optimize it anyway   I prefer 1  lt  lt  31 - 1 over   0x7fffffff because you don t need count fs  unsigned  pow  2  31     - 1 because you don t need  lt math h gt

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Just remember that it s the eighth Mersenne prime   If that s too hard  it s also the third of only four known double Mersenne primes   Edit per comment request   The Euclid-Euler theorem states that every even perfect number has the form 2  n - 1   2 n - 1   where 2 n - 1 is a prime number  The prime numbers of the form 2 n - 1 are known as Mersenne primes  and require n itself to be prime   We know that the length of an INT32 is of course 32 bits  Given the generally accepted understanding of 2 s complement  a signed INT32 is 32 bits - 1 bit    To find the magnitude of a binary number with a given number of bits we generally raise 2 to the power n  minus 1  where n is equal to the number of bits   Thus the magnitude calculation is 2  32 - 1  - 1   2 31 - 1  31 is prime and as outlined above  prime numbers of this form are Mersenne primes  We can prove it is the eight of such simply by counting them  For further details  please ask Euler  or maybe Bernoulli  to whom he wrote about them    See  https   books google ie books id x7p4tCPPuXoC amp printsec frontcover amp dq 9780883853283 amp hl en amp sa X amp ved 0ahUKEwilzbORuJLdAhUOiaYKHcsZD-EQ6AEIKTAA v onepage amp q 9780883853283 amp f false

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The most correct answer I can think of is Int32 MaxValue

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The best rule to memorize it is  21  magic number   47  just remember it  48  sequential   36  21   15  both magics   47 again  Also it is easier to remember 5 pairs than 10 digits

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The easiest way to remember is to look at  std  numeric limits lt  int  gt   max    For example  from MSDN       numeric limits max cpp   include  lt iostream gt   include  lt limits gt   using namespace std   int main        cout  lt  lt   The maximum value for type float is              lt  lt  numeric limits lt float gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type double is              lt  lt  numeric limits lt double gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type int is              lt  lt  numeric limits lt int gt   max             lt  lt  endl     cout  lt  lt   The maximum value for type short int is              lt  lt  numeric limits lt short int gt   max             lt  lt  endl

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What do you mean  It should be easy enough to remember that it is 2 32  If you want a rule to memorize the value of that number  a handy rule of thumb is for converting between binary and decimal in general   2 10   1000  which means 2 20   1 000 000  and 2 30   1 000 000 000  Double that  2 31  is rounghly 2 billion  and doubling that again  2 32  is 4 billion   It s an easy way to get a rough estimate of any binary number  10 zeroes in binary becomes 3 zeroes in decimal

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2GB   is there a minimum length for answers

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It s 2 147 483 647  Easiest way to memorize it is via a tattoo

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First write out 47 twice   you like Agent 47  right    keeping spaces as shown  each dash is a slot for a single digit  First 2 slots  then 4   --47----47   Think you have 12 in hand  because 12   a dozen   Multiply it by 4  first digit of Agent 47 s number  i e  47  and place the result to the right of first pair you already have  12   4   48 --4748--47  lt -- after placing 48 to the right of first 47   Then multiply 12 by 3  in order to make second digit of Agent 47 s number  which is 7  you need 7 - 4   3  and put the result to the right of the first 2 pairs  the last pair-slot  12   3   36 --47483647  lt -- after placing 36 to the right of first two pairs   Finally drag digits one by one from your hand starting from right-most digit  2 in this case  and place them in the first empty slot you get  2-47483647  lt -- after placing 2 2147483647  lt -- after placing 1   There you have it  For negative limit  you can think of that as 1 more in absolute value than the positive limit   Practise a few times  and you will get the hang of it

[integer] What is the maximum value for an int32?

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