How to convert an integer to a character array using C
Make use of the log10
function to determine the number of digits and do like below:
char * toArray(int number)
{
int n = log10(number) + 1;
int i;
char *numberArray = calloc(n, sizeof(char));
for (i = n-1; i >= 0; --i, number /= 10)
{
numberArray[i] = (number % 10) + '0';
}
return numberArray;
}
Or the other option is sprintf(yourCharArray,"%ld", intNumber);
'sprintf' will work fine, if your first argument is a pointer to a character (a pointer to a character is an array in 'c'), you'll have to make sure you have enough space for all the digits and a terminating '\0'. For example, If an integer uses 32 bits, it has up to 10 decimal digits. So your code should look like:
int i;
char s[11];
...
sprintf(s,"%ld", i);
Use itoa, as is shown here.
char buf[5];
// Convert 123 to string [buf]
itoa(123, buf, 10);
buf
will be a string array as you documented. You might need to increase the size of the buffer.
The easy way is by using sprintf
. I know others have suggested itoa
, but a) it isn't part of the standard library, and b) sprintf
gives you formatting options that itoa
doesn't.