How to convert integer to string in C?

Question

I tried this example:

/* itoa example */
#include <stdio.h>
#include <stdlib.h>

int main ()
{
    int i;
    char buffer [33];
    printf ("Enter a number: ");
    scanf ("%d",&i);
    itoa (i,buffer,10);
    printf ("decimal: %s\n",buffer);
    itoa (i,buffer,16);
    printf ("hexadecimal: %s\n",buffer);
    itoa (i,buffer,2);
    printf ("binary: %s\n",buffer);
    return 0;
}

but the example there doesn't work (it says the function itoa doesn't exist).

You should give more information, like the *exact* error message, what compiler you're using, and what OS. — Gabe, Mar 11 '12 at 13:13
[my-itoa](http://code.google.com/p/my-itoa/) was already suggested somewhere else on SO. — menjaraz, Mar 11 '12 at 13:46

qwertz · Accepted Answer · 2017-08-11T07:03:48.163

303

Use sprintf():

int someInt = 368;
char str[12];
sprintf(str, "%d", someInt);

All numbers that are representable by int will fit in a 12-char-array without overflow, unless your compiler is somehow using more than 32-bits for int. When using numbers with greater bitsize, e.g. long with most 64-bit compilers, you need to increase the array size—at least 21 characters for 64-bit types.

edited Aug 11 '17 at 07:03

answered Mar 11 '12 at 13:12

qwertz

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I tried running this program, and I got a runtime error. How can I get it to work properly? http://ideone.com/Xl21B4 – Anderson Green Mar 06 '13 at 02:56
@AndersonGreen The code here is correct. You typed it incorrectly. The error message tells you where. – nschum Mar 08 '13 at 09:31
`sprintf` is discussed here in more detail: http://stackoverflow.com/questions/8232634/simple-use-of-sprintf-c – Anderson Green Mar 08 '13 at 21:07
11

Often it's better to use snprintf() to cover situations where you don't know how big `str` is going to be. (eg multi-byte characters, numbers that represent counters without a limit, etc). – gone Apr 23 '14 at 09:21

score 60 · Answer 2 · edited Nov 27 '17 at 11:20

60

Making your own itoa is also easy, try this :

char* itoa(int i, char b[]){
    char const digit[] = "0123456789";
    char* p = b;
    if(i<0){
        *p++ = '-';
        i *= -1;
    }
    int shifter = i;
    do{ //Move to where representation ends
        ++p;
        shifter = shifter/10;
    }while(shifter);
    *p = '\0';
    do{ //Move back, inserting digits as u go
        *--p = digit[i%10];
        i = i/10;
    }while(i);
    return b;
}

or use the standard sprintf() function.

edited Nov 27 '17 at 11:20

Rory Mulvaney

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answered Mar 12 '12 at 02:09

bhuwansahni

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To clarify: *do while* loops are used instead of *while* for when *i* is zero. – max Feb 21 '13 at 17:34
3

There's a bug in this code: it will fail when `i = INT_MIN` because of the `i *= -1` line. – kasrak Jan 10 '15 at 19:00
1

@Max, `while` tests the condition before the loop start, `do while` tests the condition after the loop has started. – Semirix Oct 13 '15 at 22:51
@kasrak How to deal with that case. Is a larger type the only option? – user877329 Oct 02 '16 at 09:24

cnicutar · Answer 3 · 2012-03-11T14:17:14.810

50

That's because itoa isn't a standard function. Try snprintf instead.

char str[LEN];
snprintf(str, LEN, "%d", 42);

edited Mar 11 '12 at 14:17

answered Mar 11 '12 at 13:10

cnicutar

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http://www.cplusplus.com/reference/clibrary/cstdio/sprintf/ – Sebastian Flückiger Mar 11 '12 at 13:12
At least enough to hold the maximum value allowed by the integer type, I guess. – Charles Salvia Mar 11 '12 at 13:17
2

@Gabe I use `(CHAR_BIT * sizeof(int) - 1) / 3 + 2`, as [caf mentioned](http://stackoverflow.com/a/8257728/226266). – cnicutar Mar 11 '12 at 13:19
3

you don't have to `- 1` for snprintf. – Per Johansson Mar 11 '12 at 14:16
@PerJohansson That's right :-) I was thinking of `%s` for some reason. – cnicutar Mar 11 '12 at 14:18
Yeah, but `sprintf()` is better, because you don't have the unnecessary second argument. – Janko Mar 12 '12 at 15:51
3

snprintf() is safer in that you specify how much input you're taking. Otherwise, If your string has multi-byte characters, or ends up longer than you expected due to large numbers, you can overflow your buffer and crash your program (etc). – gone Apr 23 '14 at 09:06
@gone How can I ever get multibyte in that case, or is it local dependent? – user877329 Oct 02 '16 at 09:27
You need to factor multi-byte data into the calculation of the buffer size. Awkward, but what else can you do? – gone Oct 02 '16 at 13:56

How to convert integer to string in C?

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