C - Decimal to binary converting

Question

I wrote a  simple   it took me 30 minutes  program that converts decimal number to binary  I am SURE that there s a lot simpler way so can you show me  Here s the code    include  lt iostream gt   include  lt stdlib h gt   using namespace std  int a1  a2  remainder  int tab   0  int maxtab   0  int table 0   int main         system  clear        cout  lt  lt   Enter a decimal number         cin  gt  gt  a1      a2   a1    we need our number for later on so we save it in another variable      while  a1  0    dividing by two until we hit 0               remainder   a1 2    getting a remainder - decimal number 1 or 0          a1   a1 2    dividing our number by two         maxtab       1 to max elements of the table            maxtab--    -1 to max elements of the table  when dividing finishes it adds 1 additional elemnt that we don t want and it s equal to 0      a1   a2    we must do calculations one more time so we re gatting back our original number     table 0    table maxtab     we set the number of elements in our table to maxtab  we don t get 10 s of 0 s       while  a1  0    same calculations 2nd time but adding every 1 or 0  remainder  to separate element in table               remainder   a1 2    getting a remainder         a1   a1 2    dividing by 2         table tab    remainder    adding 0 or 1 to an element         tab      tab  element count  increases by 1 so next remainder is saved in another element            tab--    same as with maxtab--     cout  lt  lt   Your binary number          while  tab gt  0    until we get to the 0  1st  element of the table               cout  lt  lt  table tab   lt  lt         write the value of an element  0 or 1          tab--    decreasing by 1 so we show 0 s and 1 s FROM THE BACK  correct way             cout  lt  lt  endl      return 0      By the way it s complicated but I tried my best   edit - Here is the solution I ended up using   std  string toBinary int n        std  string r      while n  0   r  n 2  0   0   1   r  n  2       return r

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Here is modern variant that can be used for ints of different sizes    include  lt type traits gt   include  lt bitset gt   template lt typename T gt  std  enable if t lt std  is integral v lt T gt  std  string gt  encode binary T i       return std  bitset lt sizeof T    8 gt  i  to string

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A pretty straight forward solution to print binary        include  lt iostream gt  using namespace std  int main      int num arr 64    cin gt  gt num   int i 0 r   while num  0      r   num 2    arr i      r    num    2     for int j i-1 j gt  0 j--    cout lt  lt arr j

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using bitmask and bitwise and   string int2bin int n       string x      for int i 0 i lt 32 i             if n amp 1   x   1            else  x   0            n gt  gt  1            reverse x begin   x end         return x

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The conversion from natural number to a binary string   string toBinary int n        if  n  0  return  0       else if  n  1  return  1       else if  n 2    0  return toBinary n 2     0       else if  n 2    0  return toBinary n 2     1

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You can use  std  bitset to convert a number to its binary format   Use the following code snippet   std  string binary   std  bitset lt 8 gt  n  to string      I found this on stackoverflow itself  I am attaching the link

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An int variable is not in decimal  it s in binary  What you re looking for is a binary string representation of the number  which you can get by applying a mask that filters individual bits  and then printing them   for  int i   sizeof value  CHAR BIT-1  i gt  0  --i      cout  lt  lt  value  amp   1  lt  lt  i     1     0     That s the solution if your question is algorithmic  If not  you should use the std  bitset class to handle this for you   bitset lt  sizeof value  CHAR BIT  gt  bits  value    cout  lt  lt  bits to string

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std  string bin uint fast8 t i  return  i  0  i  1  1  bin i 2   i 2  1   0

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Okay   I might be a bit new to C    but I feel the above examples don t quite get the job done right   Here s my take on this situation   char  DecimalToBinary unsigned   int64 value  int bit precision        int length    bit precision   7   gt  gt  3  lt  lt  3      static char  binary   new char 1   length       int begin   length - bit precision      unsigned   int64 bit value   1      for  int n   length  --n  gt   begin                  binary n    48     value  amp  bit value     bit value           bit value  lt  lt   1            for  int n   begin  --n  gt   0            binary n    48       binary length    0      return binary       value   The Value we are checking    bit precision   The highest left most bit to check for    Length   The Maximum Byte Block Size  E g  7   1 Byte and 9   2 Byte  but we represent this in form of bits so 1 Byte   8 Bits    binary   just some dumb name I gave to call the array of chars we are setting  We set this to static so it won t be recreated with every call  For simply getting a result and display it then this works good  but if let s say you wanted to display multiple results on a UI they would all show up as the last result  This can be fixed by removing static  but make sure you delete    the results when you are done with it     begin   This is the lowest index that we are checking  Everything beyond this point is ignored  Or as shown in 2nd loop set to 0    first loop - Here we set the value to 48 and basically add a 0 or 1 to 48 based on the bool value of  value  amp  bit value     bit value  If this is true the char is set to 49  If this is false the char is set to 48  Then we shift the bit value or basically multiply it by 2    second loop - Here we set all the indexes we ignored to 48 or  0    SOME EXAMPLE OUTPUTS     int main         int val   -1      std  cout  lt  lt  DecimalToBinary val  1   lt  lt    n       std  cout  lt  lt  DecimalToBinary val  3   lt  lt    n       std  cout  lt  lt  DecimalToBinary val  7   lt  lt    n       std  cout  lt  lt  DecimalToBinary val  33   lt  lt    n       std  cout  lt  lt  DecimalToBinary val  64   lt  lt    n       std  cout  lt  lt    nPress any key to continue            std  cin ignore        return 0     00000001   Value   2 1 - 1 00000111   Value   2 3 - 1  01111111   Value   2 7 - 1  0000000111111111111111111111111111111111   Value   2 33 - 1  1111111111111111111111111111111111111111111111111111111111111111   Value   2 64 - 1    SPEED TESTS   Original Question s Answer   Method  toBinary int     Executions  10 000   Total Time  Milli   4701 15   Average Time  Nanoseconds   470114  My Version   Method  DecimalToBinary int  int       Using 64 Bit Precision   Executions  10 000 000   Total Time  Milli   3386   Average Time  Nanoseconds   338    Using 1 Bit Precision   Executions  10 000 000  Total Time  Milli   634  Average Time  Nanoseconds   63

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This is a more simple program than ever    Program to convert Decimal into Binary  include lt iostream gt  using namespace std  int main         long int dec      int rem i j bin 100  count -1      again      cout lt  lt  ENTER THE DECIMAL NUMBER -        cin gt  gt dec   input of Decimal     if dec lt 0                cout lt  lt  PLEASE ENTER A POSITIVE DECIMAL           goto again            else                   cout lt  lt   nIT s BINARY FORM IS -            for i 0 dec  0 i     making array of binary  but reversed                       rem dec 2              bin i  rem              dec dec 2              count                      for j count j gt  0 j--   reversed binary is printed in correct order                       cout lt  lt bin j                       return 0

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here a simple converter by using std  string as container  it allows a negative value    include  lt iostream gt   include  lt string gt   include  lt limits gt   int main         int x   -14       int n   std  numeric limits lt int gt   digits - 1       std  string s      s reserve n   1        do         s push back   x  gt  gt  n   amp  1     0        while --n  gt  -1        std  cout  lt  lt  s  lt  lt    n

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Non recursive solution    include  lt iostream gt   include lt string gt    std  string toBinary int n        std  string r      while n  0   r  n 2  0   0   1   r  n  2       return r    int main         std  string i  toBinary 10       std  cout lt  lt i      Recursive solution    include  lt iostream gt   include lt string gt   std  string r     std  string toBinary int n        r  n 2  0   0   1   r      if  n   2    0            toBinary n   2             return r    int main         std  string i toBinary 10       std  cout lt  lt i

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My way of converting decimal to binary in C    But since we are using mod  this function will work in case of hexadecimal or octal also  You can also specify bits  This function keeps calculating the lowest significant bit and place it on the end of the string  If you are not so similar to this method than you can vist  https   www wikihow com Convert-from-Decimal-to-Binary   include  lt bits stdc   h gt  using namespace std   string itob int bits  int n        int c      char s bits 1       1 to append NULL character       s bits      0      The NULL character in a character array flags the end of the string  not appending it may cause problems       c   bits - 1     If the length of a string is n  than the index of the last character of the string will be n - 1  Cause the index is 0 based not 1 based  Try yourself       do           if n 2  s c     1           else s c     0           n    2          c--        while  n gt 0        while c  gt  -1            s c     0           c--         return s     int main         cout  lt  lt  itob 1  0   lt  lt  endl     0 in 1 bit binary      cout  lt  lt  itob 2  1   lt  lt  endl     1 in 2 bit binary      cout  lt  lt  itob 3  2   lt  lt  endl     2 in 3 bit binary      cout  lt  lt  itob 4  4   lt  lt  endl     4 in 4 bit binary      cout  lt  lt  itob 5  15   lt  lt  endl     15 in 5 bit binary      cout  lt  lt  itob 6  30   lt  lt  endl     30 in 6 bit binary      cout  lt  lt  itob 7  61   lt  lt  endl     61 in 7 bit binary      cout  lt  lt  itob 8  127   lt  lt  endl     127 in 8 bit binary      return 0      The Output   0 01 010 0100 01111 011110 0111101 01111111

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In this approach  the decimal will be converted to the respective binary number in the string formate  The string return type is chosen since it can handle more range of input values   class Solution   public    string ConvertToBinary int num           vector lt int gt  bin      string op      for  int i   0  num  gt  0  i                bin push back num   2         num    2            reverse bin begin    bin end         for  size t i   0  i  lt  bin size      i              op    to string bin i              return op

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function to convert decimal to binary void decToBinary int n           array to store binary number     int binaryNum 1000           counter for binary array     int i   0      while  n  gt  0                storing remainder in binary array         binaryNum i    n   2          n   n   2          i                  printing binary array in reverse order     for  int j   i - 1  j  gt   0  j--          cout  lt  lt  binaryNum j       refer  - https   www geeksforgeeks org program-decimal-binary-conversion   or  using function  -   include lt bits stdc   h gt  using namespace std   int main          int n cin gt  gt n      cout lt  lt bitset lt 8 gt  n  to string   lt  lt endl        or  using left shift   include lt bits stdc   h gt  using namespace std  int main            here n is the number of bit representation we want      int n cin gt  gt n          num is a number whose binary representation we want     int num      cin gt  gt num       for int i n-1 i gt  0 i--                if  num  amp    1  lt  lt  i     cout lt  lt 1          else cout lt  lt 0

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Here are two approaches  The one is similar to your approach    include  lt iostream gt   include  lt string gt   include  lt limits gt   include  lt algorithm gt   int main         while   true                 std  cout  lt  lt   Enter a non-negative number  0-exit               unsigned long long x   0          std  cin  gt  gt  x           if    x   break           const unsigned long long base   2           std  string s          s reserve  std  numeric limits lt unsigned long long gt   digits              do   s push back  x   base    0       while   x    base             std  cout  lt  lt  std  string  s rbegin    s rend       lt  lt  std  endl            and the other uses std  bitset as others suggested    include  lt iostream gt   include  lt string gt   include  lt bitset gt   include  lt limits gt   int main         while   true                 std  cout  lt  lt   Enter a non-negative number  0-exit               unsigned long long x   0          std  cin  gt  gt  x           if    x   break           std  string s                std  bitset lt std  numeric limits lt unsigned long long gt   digits gt   x   to string             std  string  size type n   s find   1              std  cout  lt  lt  s substr  n     lt  lt  std  endl

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include  lt iostream gt   include  lt bitset gt    define bits x    std  string                std  bitset lt 8 gt  x  to string lt char std  string  traits type  std  string  allocator type gt      c str       int main         std  cout  lt  lt  bits  -86  gt  gt  1    lt  lt        lt  lt   -86  gt  gt  1   lt  lt  std  endl      return 0

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include  lt iostream gt      x is our number to test    pow is a power of 2  e g  128  64  32  etc     int printandDecrementBit int x  int pow           Test whether our x is greater than some power of 2 and print the bit     if  x  gt   pow                std  cout  lt  lt   1              If x is greater than our power of 2  subtract the power of 2         return x - pow            else               std  cout  lt  lt   0           return x           int main         std  cout  lt  lt   Enter an integer between 0 and 255         int x      std  cin  gt  gt  x       x   printandDecrementBit x  128       x   printandDecrementBit x  64       x   printandDecrementBit x  32       x   printandDecrementBit x  16        std  cout  lt  lt            x   printandDecrementBit x  8       x   printandDecrementBit x  4       x   printandDecrementBit x  2       x   printandDecrementBit x  1        return 0      this is a simple way to get the binary form of an int  credit to learncpp com  im sure this could be used in different ways to get to the same point

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Below is simple C code that converts binary to decimal and back again   I wrote it long ago for a project in which the target was an embedded processor and the development tools had a stdlib that was way too big for the firmware ROM  This is generic C code that does not use any library  nor does it use division or the remainder     operator  which is slow on some embedded processors   nor does it use any floating point  nor does it use any table lookup nor emulate any BCD arithmetic   What it does make use of is the type long long  more specifically unsigned long long  or uint64 t   so if your embedded processor  and the C compiler that goes with it  cannot do 64-bit integer arithmetic  this code is not for your application   Otherwise  I think this is production quality C code  maybe after changing long to int32 t and unsigned long long to uint64 t    I have run this overnight to test it for every 2     signed integer values and there is no error in conversion in either direction  We had a C compiler linker that could generate executables and we needed to do what we could do without any stdlib  which was a pig    So no printf   nor scanf     Not even an sprintf   nor sscanf     But we still had a user interface and had to convert base-10 numbers into binary and back    We also made up our own malloc  -like utility also and our own transcendental math functions too   So this was how I did it  the main program and calls to stdlib were there for testing this thing on my mac  not for the embedded code    Also  because some older dev systems don t recognize  quot int64 t quot  and  quot uint64 t quot  and similar types  the types long long and unsigned long long are used and assumed to be the same   And long is assumed to be 32 bits   I guess I could have typedefed it     returns an error code  0 if no error     -1 if too big  -2 for other formatting errors int decimal to binary char  dec  long  bin            int i   0           int past leading space   0      while  i  lt   64  amp  amp   past leading space            first get past leading spaces                   if  dec i                                    i                           else                           past leading space   1                              if   past leading space                    return -2                                    64 leading spaces does not a number make                  at this point the only legitimate remaining        chars are decimal digits or a leading plus or minus sign      int negative   0      if  dec i      -                     negative   1          i                   else if  dec i                            i                                          do nothing but go on to next char                  now the only legitimate chars are decimal digits     if  dec i       0                     return -2                                  there needs to be at least one good                                                     digit before terminating string          unsigned long abs bin   0      while  i  lt   64  amp  amp  dec i       0                     if   dec i   gt    0   amp  amp  dec i   lt    9                              if  abs bin  gt  214748364                                    return -1                                    this is going to be too big                               abs bin    10                                    previous value gets bumped to the left one digit                                abs bin     unsigned long  dec i  -  0               and a new digit appended to the right             i                           else                           return -2                                        not a legit digit in text string                                  if  dec i       0                     return -2                                    not terminated string in 64 chars                    if  negative                    if  abs bin  gt  2147483648                            return -1                                too big                        bin   - long abs bin                 else                   if  abs bin  gt  2147483647                            return -1                                too big                        bin    long abs bin                     return 0          void binary to decimal char  dec  long bin            unsigned long long acc                    64-bit unsigned integer          if  bin  lt  0                      dec       -                         leading minus sign         bin   -bin                            make bin value positive                    acc   989312855LL  unsigned long bin            very nearly 0 2303423488   2 32     acc    0x00000000FFFFFFFFLL                     we need to round up     acc  gt  gt   32      acc    57646075LL  unsigned long bin          2 59   10 10      57646075 2303423488     57646075    989312854 979825   2 32             int past leading zeros   0      for  int i 9  i gt  0  i--                maximum number of digits is 10                   acc  lt  lt   1          acc     acc lt  lt 2                     an efficient way to multiply a long long by 10         acc    10                   unsigned int digit    unsigned int  acc  gt  gt  59             the digit we want is in bits 59 - 62                  if  digit  gt  0                            past leading zeros   1                                 if  past leading zeros                              dec       0    digit                                 acc  amp   0x07FFFFFFFFFFFFFFLL        mask off this digit and go on to the next digit                    if   past leading zeros                if all digits are zero                         dec       0                          put in at least one zero digit                     dec     0                             terminate string          if 1   include  lt stdlib h gt   include  lt stdio h gt  int main  int argc  const char  argv              char dec 64       long bin  result1  result2      unsigned long num errors      long long long long bin           num errors   0      for  long long bin -2147483648LL  long long bin lt  2147483647LL  long long bin                      bin    long long long bin          if   bin amp 0x00FFFFFFL     0                            printf  quot bin    ld  n quot   bin             this is to tell us that things are moving along                       binary to decimal dec  bin           decimal to binary dec   amp result1           sscanf dec   quot  ld quot    amp result2                 decimal to binary   should do the same as this sscanf                    if  bin    result1    bin    result2                            num errors                printf  quot bin    ld  result1    ld  result2    ld  num errors    ld  dec    s  n quot                   bin  result1  result2  num errors  dec                                    printf  quot num errors    ld  n quot   num errors            return 0          else   include  lt stdlib h gt   include  lt stdio h gt  int main  int argc  const char  argv              char dec 64       long bin           printf  quot bin    quot        scanf  quot  ld quot    amp bin       while  bin    0                    binary to decimal dec  bin           printf  quot dec    s  n quot   dec           printf  quot bin    quot            scanf  quot  ld quot    amp bin                      return 0          endif

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DECIMAL TO BINARY NO ARRAYS USED  made by Oya   I m still a beginner  so this code will only use loops and variables xD     Hope you like it  This can probably be made simpler than it is          include  lt iostream gt       include  lt cmath gt       include  lt cstdlib gt       using namespace std       int main                 int i          int expoentes    the sequence  gt  pow 2 i  or 2 i         int decimal           int extra    this will be used to add some 0s between the 1s         int x   1           cout  lt  lt    nThis program converts natural numbers into binary code nPlease enter a Natural number            cout  lt  lt    n nWARNING  Only works until  1 073 millions n           cout  lt  lt        To exit  enter a negative number n n            while decimal  gt   0               cout  lt  lt    n-----    ----- n n               cin  gt  gt  decimal              cout  lt  lt    n                if decimal    0                   cout  lt  lt   0                             while decimal  gt   1                   i   0                  expoentes   1                  while decimal  gt   expoentes                       i                        expoentes   pow 2 i                                     x   1                  cout  lt  lt   1                   decimal -  pow 2 i-x                   extra   pow 2 i-1-x                   while decimal  lt  extra                       cout  lt  lt   0                       x                        extra   pow 2 i-1-x                                                     return 0

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You want to do something like   cout  lt  lt   Enter a decimal number     cin  gt  gt  a1  cout  lt  lt  setbase 2   cout  lt  lt  a1

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include  lt stdio h gt   include  lt stdlib h gt   include  lt string h gt   include  lt math h gt   void Decimal2Binary long value char  b int len        if value gt 0                do                       if value  1                                  b len-1   1                   break                            else                                 b len-1   value 2  48                  value value 2                  len--                         while 1           long Binary2Decimal char  b int len        int i 0      int j 0      long value 0      for i  len-1  i gt  0 i--                if   b i   49                        value  pow 2 j                     j              return value    int main         char data 11        BIT             long value 1023      memset data  0  sizeof data        data 10    0             Decimal2Binary value data 10       printf   d- gt  s n  value data       value Binary2Decimal data 10       printf   s- gt  d  data value       return 0

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The following is a recursive function which takes a positive integer and prints its binary digits to the console   Alex suggested  for efficiency  you may want to remove printf   and store the result in memory    depending on storage method result may be reversed          Takes a unsigned integer  converts it into binary and prints it to the console      param n the number to convert and print     void convertToBinary unsigned int n        if  n   2    0            convertToBinary n   2             printf   d   n   2       Credits to UoA ENGGEN 131   Note  The benefit of using an unsigned int is that it can t be negative

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For this   In C   you can use itoa   function  This function convert any Decimal integer to binary  decimal   hexadecimal and octal number    include lt bits stdc   h gt  using namespace std  int main     int a       char res 1000    cin gt  gt a   itoa a res 10    cout lt  lt  Decimal-   lt  lt res lt  lt endl   itoa a res 2    cout lt  lt  Binary-   lt  lt res lt  lt endl   itoa a res 16    cout lt  lt  Hexadecimal-   lt  lt res lt  lt endl   itoa a res 8    cout lt  lt  Octal-   lt  lt res lt  lt endl return 0      However  it is only supported by specific compilers   You can see also  itoa - C   Reference

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std  bitset has a  to string   method that returns a std  string holding a text representation in binary  with leading-zero padding   Choose the width of the bitset as needed for your data  e g  std  bitset lt 32 gt  to get 32-character strings from 32-bit integers    include  lt iostream gt   include  lt bitset gt   int main         std  string binary   std  bitset lt 8 gt  128  to string      to binary     std  cout lt  lt binary lt  lt   n        unsigned long decimal   std  bitset lt 8 gt  binary  to ulong        std  cout lt  lt decimal lt  lt   n       return 0      EDIT  Please do not edit my answer for Octal and Hexadecimal  The OP specifically asked for Decimal To Binary

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include  stdafx h   include lt iostream gt   include lt vector gt   include lt cmath gt   using namespace std   int main            Initialize Variables     double x      int xOct      int xHex         Initialize a variable that stores the order if the numbers in binary sexagesimal base     vector lt int gt  rem         Get Demical value     cout  lt  lt   Number  demical base          cin  gt  gt  x         Set the variables     xOct   x      xHex   x         Get the binary value     for  int i   0  x  gt   1  i              rem push back abs remainder x  2             x   floor x   2                Print binary value     cout  lt  lt   Binary         int n   rem size        while  n  gt  0            n--          cout  lt  lt  rem n         cout  lt  lt  endl         Print octal base     cout  lt  lt  oct  lt  lt   Octal     lt  lt  xOct  lt  lt  endl         Print hexademical base     cout  lt  lt  hex  lt  lt   Hexademical     lt  lt  xHex  lt  lt  endl       system  pause        return 0

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HOPE YOU LIKE THIS SIMPLE CODE OF CONVERSION FROM DECIMAL TO BINARY      include lt iostream gt      using namespace std      int main                 int input rem res count 0 i 0          cout lt  lt  Input number             cin gt  gt input  enter code here          int num input          while input  gt  0                        input input 2                count                       int arr count            while num  gt  0                        arr i  num 2              num num 2                i                      for int i count-1   i gt  0   i--                        cout lt  lt      lt  lt  arr i  lt  lt                           return 0

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Your solution needs a modification  The final string should be reversed before returning   std  reverse r begin    r end     return r

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include  lt iostream gt  using namespace std   int main           int a b      cin gt  gt a      for int i 31 i gt  0 i--                b  a gt  gt i  amp 1          cout lt  lt b

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There is in fact a very simple way to do so  What we do is using a recursive function which is given the number  int  in the parameter  It is pretty easy to understand  You can add other conditions variations too  Here is the code   int binary int num        int rem      if  num  lt   1                        cout  lt  lt  num              return num                rem   num   2      binary num   2       cout  lt  lt  rem      return rem

[c++] C++ - Decimal to binary converting

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