'\0' evaluates false, "\0" evaluates true

c arrays string null kernighan-and-ritchie
Rahn picture Rahn · Apr 30, 2016 · Viewed 8.9k times · Source

Inspired by a program described in K&R section 5.5:

void strcpy(char *s, char *t)
{
    while(*s++ = *t++);
}

C program

if ('\0') { printf("\'\\0\' -> true \n"); }
else      { printf("\'\\0\' -> false\n"); }

if ("\0") { printf("\"\\0\" -> true \n"); }
else      { printf("\"\\0\" -> false\n"); }

prints

'\0' -> false
"\0" -> true

Why do '\0' and "\0" evaluate differently in C?

clang version 3.8.0

Answer

user253751 picture user253751 · Apr 30, 2016

Recall how string literals work in C - "\0" is a character array containing two zero bytes (the one you asked for, and the implicit one at the end). When evaluated for the if test, it decays into a pointer to its first character. This pointer is not NULL, so it's considered true when used as a condition.

'\0' is the number zero, equivalent to just 0. It's an integer which is zero, so it's considered false when used as a condition.

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