I'm trying to print all characters stored in hex array to the screen, one by one, but I get this strange error in line 16. As far as I know, %c should be expecting a char, not an int.
Why I'm getting this error?
Below is my code, thanks.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
int main()
{
char hex[8] = "cf0a441f";
int hexCounter;
char *currentHex;
for(hexCounter=0; hexCounter<strlen(hex); hexCounter++)
{
currentHex = &hex[hexCounter];
printf("%c",currentHex);
}
return 0;
}
Best Answer
You mean
In my opinion you can remove the entire idea of
currentHex
since it just adds complexity for no value. Simply do:The important point is that you're supposed to pass the value of the character itself, not it's address which is what you're doing.