Identical string literals are considered equal? [duplicate]

Answer 1

It's unspecified for string literals with the same content to have the same address or not. So the output of your program could be equal or it could be not equal, your compiler happens to put them in the same place.

C11 6.4.5 String literals

It is unspecified whether these arrays are distinct provided their elements have the appropriate values.

Answer 2

Of course, what you do in that condition is a comparison between pointers (use strcmp to compare C strings).

So, I think it is a compiler translation/optimization that "maps" identical literals at the same location in the memory.

EDIT 1:

The following sample confirms what I wrote:

#include <stdio.h>

char* a = "ffffd";
char* b = "ffffd";
char* c = "ffffd";

int main() {
    printf ("a => %p\nb => %p\nc => %p\n", a, b, c);
}

The previous program, compiled with gcc using -O0 and executed will print:

a => 0x40060c
b => 0x40060c
c => 0x40060c

I don't know how other compilers will treat the same situation.

Answer 3

When you're comparing two character values (which are not pointers), it is a numeric comparison.

But when you're comparing two string, Base address of strings are compared.If supposed compilers treats as both string are in same location ,then o/p is equal.Otherwise Not.

What you are comparing the two memory addresses for the different strings, which are stored in different locations.so,Not equal.

Even it is read only memory,you are used this for comparison only.You are not modify or not write anything.