Convert integer into an array

Question

Hi,

I am working on a C program, and I am coming across a small problem. I don't know how to convert an integer (say 2007) into a char array. Is there a function in the C libraries to do that for me?

To clarify, I'd like to take 2007 and store it in some char array[4] ={ '2', '0', '0', '7', '\0' };

I was thinking something like sprintf, but I'm not sure. Anyways, any help/hints would be appreciated.

Thanks, Michael

Answer 1

Use snprintf:

// value is int;
// buf is char *;
// length is space available in buf
snprintf(buf, length, "%d", value)

snprintf has the advantage of being standard and giving your more flexibility over the formatting as well as being safe.

You could also use itoa but be warned that it is not part of the standard. Most implementations have it though.

Usage:

// value is int;
// buf is char *;
itoa(value, buf, 10);

An interesting question is: how much space do you allocate for buf? We note the following. With sizeof(int) bytes per int and eight bits per byte, the maximum value is approximately 2^(CHAR_BIT * sizeof(int) - 1) (-1 is for sign bit). Therefore we need space to hold

floor(log_10(2^(CHAR_BIT * sizeof(int) - 1)) + 1

digits. But don't forget the sign and null terminator! So the maximum length of an integer representable here is

floor(log_10(2^(CHAR_BIT * sizeof(int) - 1)) + 3.

Of course, this could be wasting space if our values are small. To find out how much space a specific value needs:

floor(log_10(abs(value))) + 1 + (value < 0 ? 1 : 0) + 1

Answer 2

The classic way is itoa. Or you can use snprintf to get more control.

Answer 3

You can do that with sprintf, or more safely snprintf.

Here's a contrived example:

#include <stdio.h>
#define MAX_LEN 5

char str[MAX_LEN];
snprintf(str, MAX_LEN, "%d", 2007);

Answer 4

There is a nonstandard, but well-supported as I undertand it, function itoa - the opposite of atoi.

Example:

char *a = malloc(10 * sizeof(char));
itoa(2007, a, 2007);

sprintf also works:

char *a = malloc(10 * sizeof(char));
sprintf(a, "%d", 2007);

Answer 5

#include <stdio.h>
char array[5];
sprintf(array, "%d", 2007);

...done.

Note that sprintf is not overflow safe, so if you have a number of 5 or more digits you'll have a problem. Note also that the converted number will be followed by a terminating \0.

Answer 6

int n = 2007;
char a[100];
sprintf( a, "%d", n );

Answer 7

Use snprintf() and be sure allocate the proper amount of space to hold numbers up to 2^(sizeof(int)*CHAR_BIT). On a 64-bit machine, that will be 20 digit characters, plus 1 for the NULL terminator.

#include <stdio.h>
#define MAX_DIGITS 20

int n = 2007;   
char str[MAX_DIGITS+1];
snprintf(str, MAX_DIGITS+1, "%d", n);