"Saif" <saifch@gmail.com> writes:
[color=blue]
>
henrytcy@gmail.com wrote:[color=green]
> > How can I convert integer, for example 12113, to char array but without
> > specify an array size in C programming?[/color]
>
> You can use dynamically allocated memory. To get the size of the string
> needed, you can find the log (base 10) of your number, truncate it, and
> add 1.
>
> For example for 12113...
>
> size = floor(log10(12113)) + 1 = 4 + 1 = 5[/color]
If I'm going to convert it to an char array I'd use snprintf to
calculate the size:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char* buf = NULL;
int x = 12113;
size_t sz = snprintf(buf, 0, "%d", x) + 1;
buf = malloc(sz);
snprintf(buf, sz, "%d", x);
/* do something with buf */
free(buf);
return 0;
}
