biggest integer that can be stored in a double

Question

What is the biggest  no-floating  integer that can be stored in an IEEE 754 double type without losing precision

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You need to look at the size of the mantissa   An IEEE 754 64 bit floating point number  which has 52 bits  plus 1 implied  can exactly represent integers with an absolute value of less than or equal to 2 53

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The largest integer that can be represented in IEEE 754 double  64-bit  is the same as the largest value that the type can represent  since that value is itself an integer   This is represented as 0x7FEFFFFFFFFFFFFF  which is made up of    The sign bit 0  positive  rather than 1  negative  The maximum exponent 0x7FE  2046 which represents 1023 after the bias is subtracted  rather than 0x7FF  2047 which indicates a NaN or infinity   The maximum mantissa 0xFFFFFFFFFFFFF which is 52 bits all 1    In binary  the value is the implicit 1 followed by another 52 ones from the mantissa  then 971 zeros  1023 - 52   971  from the exponent   The exact decimal value is   179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368  This is approximately 1 8 x 10308

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The biggest largest integer that can be stored in a double without losing precision is the same as the largest possible value of a double  That is  DBL MAX or approximately 1 8 nbsp    nbsp 10308  if your double is an IEEE 754 64-bit double   It s an integer  It s represented exactly  What more do you want   Go on  ask me what the largest integer is  such that it and all smaller integers can be stored in IEEE 64-bit doubles without losing precision  An IEEE 64-bit double has 52 bits of mantissa  so I think it s 253    253   1 cannot be stored  because the 1 at the start and the 1 at the end have too many zeros in between   Anything less than 253 can be stored  with 52 bits explicitly stored in the mantissa  and then the exponent in effect giving you another one   253 obviously can be stored  since it s a small power of 2    Or another way of looking at it  once the bias has been taken off the exponent  and ignoring the sign bit as irrelevant to the question  the value stored by a double is a power of 2  plus a 52-bit integer multiplied by 2exponent nbsp - nbsp 52  So with exponent 52 you can store all values from 252 through to 253 nbsp - nbsp 1  Then with exponent 53  the next number you can store after 253 is 253   1 nbsp    nbsp 253 - 52  So loss of precision first occurs with 253   1

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DECIMAL DIG from  lt float h gt  should give at least a reasonable approximation of that  Since that deals with decimal digits  and it s really stored in binary  you can probably store something a little larger without losing precision  but exactly how much is hard to say  I suppose you should be able to figure it out from FLT RADIX and DBL MANT DIG  but I m not sure I d completely trust the result

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9007199254740992  that s 9 007 199 254 740 992  with no guarantees     Program   include  lt math h gt   include  lt stdio h gt   int main void      double dbl   0     I started with 9007199254000000  a little less than 2 53      while  dbl   1    dbl  dbl      printf    0f n   dbl - 1     printf    0f n   dbl     printf    0f n   dbl   1     return 0      Result   9007199254740991 9007199254740992 9007199254740992

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It is true that  for 64-bit IEEE754 double  all integers up to 9007199254740992    2 53 can be exactly represented  However  it is also worth mentioning that all representable numbers beyond 4503599627370496    2 52 are integers  Beyond 2 52 it becomes meaningless to test whether or not they are integers  because they are all implicitly rounded to a nearby representable value  In the range 2 51 to 2 52  the only non-integer values are the midpoints ending with  quot  5 quot   meaning any integer test after a calculation must be expected to yield at least 50  false answers  Below 2 51 we also have  quot  25 quot  and  quot  75 quot   so comparing a number with its rounded counterpart in order to determine if it may be integer or not starts making some sense  TLDR  If you want to test whether a calculated result may be integer  avoid numbers larger than 2251799813685248    2 51

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1 7976931348623157    10 308   http   en wikipedia org wiki Double precision floating-point format

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Wikipedia has this to say in the same context with a link to IEEE 754      On a typical computer system  a  double precision   64-bit  binary floating-point number has a coefficient of 53 bits  one of which is implied   an exponent of 11 bits  and one sign bit    2 53 is just over 9   10 15

[types] biggest integer that can be stored in a double

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