What does empty string contain in C++

Question

Following code

#include <stdio.h> int main( int argc, char** argv ) { const char *s = ""; if (s == '\0') { int x = 0; } return 0; }

It does not go in the loop. Why ? ,

You're comparing the address of a string with a character (promoted to an integer), but yes, "" is just one null character. — Qaz
– Qaz, Commented Jan 28, 2013 at 5:30
It does not go in the loop. Why ? There is no loop. That's like asking a homeless man why he doesn't go home. — ta.speot.is
– ta.speot.is, Commented Jan 28, 2013 at 5:33

Jerry Coffin · Accepted Answer · 2013-01-28 05:32:19Z

You've defined s as a pointer to char. As it happens, '\0' is an integer constant expression with the value 0 -- the definition of a null pointer constant.

IOW, you're doing the equivalent of if (s == NULL). Since s actually points at a string literal, it's not a null pointer, so the comparison is false.

I'd guess what you intended is if (*s == '\0') ..., which should compare as true.

Oleksi · Accepted Answer · 2013-01-28 05:31:45Z

Try

 if (*s == '\0') { int x = 0; }

You want to compare the value of s, not it's memory address.

billz · Accepted Answer · 2013-01-28 05:40:04Z

s is a pointer, this version compares pointer

const char *s = null; if (s == '\0') { int x = 0; } return 0;

and this version compares first element of string to detect a null string:

if (s[0] == '\0') { int x = 0; }

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