Question

I was playing around with unicode characters (without using wchar_t support) just for fun. I’m only using the regular char data type. I noticed that while printing them in hex they were showing up full 4 bytes instead of just one byte.

For ex. consider this c file:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    char *s = (char *) malloc(100);
    fgets(s, 100, stdin);
    while (s && *s != '\0') {
            printf("%x\n", *s);
            s++;
    }
    return 0;
}

After compiling with gcc and giving input as ‘cent’ symbol (hex: c2 a2) I get the following output

$ ./a.out
¢
ffffffc2: ?
ffffffa2: ?
a: 

So instead of just printing c2 and a2 I got the whole 4 bytes as if it’s an int type.

Does this mean char is not really 1-byte in length, ascii made it look like 1-byte?

Answer 1

No. printf is a variable argument function, arguments to a variable argument function will be promoted to an int. And in this case the char was negative, so it gets sign extended.