Question

In general I know why you get this error, but I’m a little confused in this particular instance…

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

char* mystrncpy(char* dst, const char* src, size_t n) {
    char* temp = dst;

    while (n-- > 0 && (*temp = *src)) {
        temp++;
        src++;
    }

    return dst;
}

int main() {
    const char* str = "Hello World!";
    char buf[50];
    memset(buf, 0, sizeof(buf));
    mystrncpy(buf, str, sizeof(buf));
    printf("%s\n", buf);
}

The code above works great, but if I remove the extra set of brackets changing the while loop to:

while (n-- > 0 && *temp = *src)

Then I get the error. I guess it’s something about operator precedence but I’m a little baffled. Can someone explain what that while loop alteration does to make this compiler error appear?

Answer 1

Yes, it’s about precedence.

= has lower precedence than &&. Therefore, the latter case parses as:

while ((n-- > 0 && *temp) = *src)

And of course (n-- > 0 && *temp) is not an l-value.

Are you sure you didn’t intend to use == instead of =?