How to convert an int to string in C?

Solution 1:

You can use sprintf to do it, or maybe snprintf if you have it:

char str[ENOUGH];
sprintf(str, "%d", 42);

Where the number of characters (plus terminating char) in the str can be calculated using:

(int)((ceil(log10(num))+1)*sizeof(char))

Solution 2:

EDIT: As pointed out in the comment, itoa() is not a standard, so better use sprintf() approach suggested in the rivaling answer!

You can use itoa() function to convert your integer value to a string.

Here is an example:

int num = 321;
char snum[5];

// convert 123 to string [buf]
itoa(num, snum, 10);

// print our string
printf("%s\n", snum);

If you want to output your structure into a file there is no need to convert any value beforehand. You can just use the printf format specification to indicate how to output your values and use any of the operators from printf family to output your data.

Solution 3:

The short answer is:

snprintf( str, size, "%d", x );

The longer is: first you need to find out sufficient size. snprintf tells you length if you call it with NULL, 0 as first parameters:

snprintf( NULL, 0, "%d", x );

Allocate one character more for null-terminator.

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

int x = -42;
int length = snprintf( NULL, 0, "%d", x );
char* str = malloc( length + 1 );
snprintf( str, length + 1, "%d", x );
...
free(str);

If works for every format string, so you can convert float or double to string by using "%g", you can convert int to hex using "%x", and so on.

Solution 4:

After having looked at various versions of itoa for gcc, the most flexible version I have found that is capable of handling conversions to binary, decimal and hexadecimal, both positive and negative is the fourth version found at http://www.strudel.org.uk/itoa/. While sprintf/snprintf have advantages, they will not handle negative numbers for anything other than decimal conversion. Since the link above is either off-line or no longer active, I've included their 4th version below:

/**
 * C++ version 0.4 char* style "itoa":
 * Written by Lukás Chmela
 * Released under GPLv3.
 */
char* itoa(int value, char* result, int base) {
    // check that the base if valid
    if (base < 2 || base > 36) { *result = '\0'; return result; }

    char* ptr = result, *ptr1 = result, tmp_char;
    int tmp_value;

    do {
        tmp_value = value;
        value /= base;
        *ptr++ = "zyxwvutsrqponmlkjihgfedcba9876543210123456789abcdefghijklmnopqrstuvwxyz" [35 + (tmp_value - value * base)];
    } while ( value );

    // Apply negative sign
    if (tmp_value < 0) *ptr++ = '-';
    *ptr-- = '\0';
    while(ptr1 < ptr) {
        tmp_char = *ptr;
        *ptr--= *ptr1;
        *ptr1++ = tmp_char;
    }
    return result;
}

Solution 5:

This is old but here's another way.

#include <stdio.h>

#define atoa(x) #x

int main(int argc, char *argv[])
{
    char *string = atoa(1234567890);
    printf("%s\n", string);
    return 0;
}