Converting integer to string in C without sprintf

Question

It is possible to convert integer to string in C without sprintf?

Answer 1

There's a nonstandard function:

char *string = itoa(numberToConvert, 10); // assuming you want a base-10 representation

Edit: it seems you want some algorithm to do this. Here's how in base-10:

#include <stdio.h>

#define STRINGIFY(x) #x
#define INTMIN_STR STRINGIFY(INT_MIN)

int main() {
    int anInteger = -13765; // or whatever

    if (anInteger == INT_MIN) { // handle corner case
        puts(INTMIN_STR);
        return 0;
    }

    int flag = 0;
    char str[128] = { 0 }; // large enough for an int even on 64-bit
    int i = 126;
    if (anInteger < 0) {
        flag = 1;
        anInteger = -anInteger;
    }

    while (anInteger != 0) { 
        str[i--] = (anInteger % 10) + '0';
        anInteger /= 10;
    }

    if (flag) str[i--] = '-';

    printf("The number was: %s\n", str + i + 1);

    return 0;
}

Answer 2

Here's an example of how it might work. Given a buffer and a size, we'll keep dividing by 10 and fill the buffer with digits. We'll return -1 if there is not enough space in the buffer.

int
integer_to_string(char *buf, size_t bufsize, int n)
{
   char *start;

   // Handle negative numbers.
   //
   if (n < 0)
   {
      if (!bufsize)
         return -1;

      *buf++ = '-';
      bufsize--;
   }

   // Remember the start of the string...  This will come into play
   // at the end.
   //
   start = buf;

   do
   {
      // Handle the current digit.
      //
      int digit;
      if (!bufsize)
         return -1;
      digit = n % 10;
      if (digit < 0)
         digit *= -1;
      *buf++ = digit + '0';
      bufsize--;
      n /= 10;
   } while (n);

   // Terminate the string.
   //
   if (!bufsize)
      return -1;
   *buf = 0;

   // We wrote the string backwards, i.e. with least significant digits first.
   // Now reverse the string.
   //
   --buf;
   while (start < buf)
   {
      char a = *start;
      *start = *buf;
      *buf = a;
      ++start;
      --buf;
   }

   return 0;
}

Answer 3

Unfortunately none of the answers above can really work out in a clean way in a situation where you need to concoct a string of alphanumeric characters.There are really weird cases I've seen, especially in interviews and at work.

The only bad part of the code is that you need to know the bounds of the integer so you can allocate "string" properly.

In spite of C being hailed predictable, it can have weird behaviour in a large system if you get lost in the coding.

The solution below returns a string of the integer representation with a null terminating character. This does not rely on any outer functions and works on negative integers as well!!

#include <stdio.h>
#include <stdlib.h>


void IntegertoString(char * string, int number) {

   if(number == 0) { string[0] = '0'; return; };
   int divide = 0;
   int modResult;
   int  length = 0;
   int isNegative = 0;
   int  copyOfNumber;
   int offset = 0;
   copyOfNumber = number;
   if( number < 0 ) {
     isNegative = 1;
     number = 0 - number;
     length++;
   }
   while(copyOfNumber != 0) 
   { 
     length++;
     copyOfNumber /= 10;
   }

   for(divide = 0; divide < length; divide++) {
     modResult = number % 10;
     number    = number / 10;
     string[length - (divide + 1)] = modResult + '0';
   }
   if(isNegative) { 
   string[0] = '-';
   }
   string[length] = '\0';
}

int main(void) {


  char string[10];
  int number = -131230;
  IntegertoString(string, number);
  printf("%s\n", string);

  return 0;
}

Answer 4

    int i = 24344; /*integer*/
    char *str = itoa(i); 
    /*allocates required memory and 
    then converts integer to string and the address of first byte of memory is returned to str pointer.*/