Comparing string with if condition

Question

I am comparing string using if condition but the second if condition is failing. What will be the reason of fail and what happens in compiler level in these comparing cases?

char *c = "test";
char a[] = "test";

if(c=="test")
{
    printf("Hai\n");
    if(a=="test")
    printf("Bye\n");
}

You really should follow a tutorial. `strcmp` is what you need. — Fiddling Bits, Jan 27 '20 at 14:54
`c=="test"` is comparing two pointers, which will (probably) always be equal in your case. `a=="test"` also compares two pointers, but those will never be equal. — Some programmer dude, Jan 27 '20 at 14:56
Or better yet, read a [good book.](https://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list). — Blastfurnace, Jan 27 '20 at 14:56
so instead of `if(c == "test")` use `if(!strcmp(c, "test"))` — anton-tchekov, Jan 27 '20 at 18:24

anastaciu · Answer 1 · 2020-01-27T16:14:35.353

In the first comparison you should get a warning:

warning: result of comparison against a string literal is unspecified

However it returns true because you are comparing two pointers to string literal, witch is always, as far as I can tell, the same. Take the following code:

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

int main(void) {
  char *c = "test";
  char a[] = "test";

  printf("%p\n%p\n%p", c, a, "test");
}

The results will be:

0x4005f4

0x7ffedb3caaf3

0x4005f4

As you can see the pointers are indeed the same.

That said, == is not used in C to compare strings, you should use strcmp().

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

int main(void) {
  char *c = "test";
  char a[] = "test";

  if(!strcmp(c, "test"))
  {
    printf("Hai\n");
    if(!strcmp(a, "test"))
      printf("Bye\n");
  }
}

I would change `c == "test" is not used in C` to `== is not used in C to compare strings`. — Fiddling Bits, Jan 27 '20 at 14:59

Comparing string with if condition

1 Answers1