Question

The following program does not work as I intended.

#include <string.h>
#include <stdio.h>

int main()
{
    const char *c = "abcdef";

    // prints 'f' as expected
    printf("%c\n", c[5]);

    // comparison fails, does not print "yes"
    if (c[5] == 'f')
        printf("yes");
    return 0;
}

How can I compare a character in a string literal to a character value? Is it possible without using ASCII related functions i.e. chr() or ord() suppose those exist. ( I remember using them back in Pascal days)

Thanks

Answer 1

There must be something else going on with your environment. I just threw that into gcc and it prints out “yes” as expected.

Are you sure you aren’t just missing the “yes” because printf does not add a new line by default, so it could be “yes[your shell prompt]” and you’re just overlooking the yes?