Create char array from int in C

Question

How do you put an int into a char array?

int x = 21, i = 3;
char length[4];

while(i >= 0) {
    length[i] = (char) (x % 10);
    x /= 10;
    i--;
} printf("%s\n", length);

// length should now be "0021"

The string comes out blank instead.

Note: This is not a duplicate of "How do I convert from int to chars in C++?" because I also need padding. i.e. "0021" not "21"

what are you trying to do? -> snprintf(length, sizeof(length), "%d", x);? — Iharob Al Asimi
– Iharob Al Asimi, Commented Mar 27, 2015 at 12:37
possible duplicate of How do I convert from int to chars in C++? — MuertoExcobito
– MuertoExcobito, Commented Mar 27, 2015 at 12:39
sprintf(length, "%04d", x); Of course, you'll want to increase the size of length, for the null terminator. — MuertoExcobito
– MuertoExcobito, Commented Mar 27, 2015 at 12:53
@RussellHickey you can add padding like this snprintf(length, sizeof(length), "%0*d", numberOfDigits, value); — Iharob Al Asimi
– Iharob Al Asimi, Commented Mar 27, 2015 at 12:53

Barmar · Accepted Answer · 2015-03-27 12:40:03Z

6

You're not getting the character code of the digit, you're using the digit as if it were its own character code. It should be:

length[i] = '0' + (x % 10);

You also need to add an extra element to the length array for the terminating null character:

char length[5];
length[4] = 0;

answered Mar 27, 2015 at 12:40

Barmar

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3 Comments

Dobz Over a year ago

Thank You, it worked. I'll accept the answer when allowed. :)

Lawrence Aiello Over a year ago

Shouldn't it be length[4] = '\0';?

Barmar Over a year ago

They're the same thing.

Iharob Al Asimi · Accepted Answer · 2015-03-27 13:29:27Z

The problem in your code is basically that 1 != '1' i.e. the character is not the integer, you need to check the ascii table to see what ascii code represents the character '1' but you don't really need to know the number, you can just use '1' note the single qoutes.

But you also didn't nul terminate your string, you need to add a '\0' at the end of the string, so

int x = 21, i = 3;
char length[5];

length[4] = '\0';
while (i >= 0) 
{
    length[i--] = x % 10 + '0';
    x          /= 10;
} 
printf("%s\n", length);

should work, but is unecessary, you can just

snprintf(length, sizeof(length), "%0*d", padding, x);
/*                                         ^ this is how many characters you want */

notice that sizeof works because length is a char array, do not confuse that with the length of a string.

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