Why first code output is false and true in second code?

Question

Code 1

#include<stdio.h>    

int main(){
const char st1[]={"Hello"};
const char st2[]={"Hello"};

if(st1==st2){
  printf("True");
}
else{
  printf("False");
}
return 0;
}

Code 2

int main(){
const char *st1="Hello";
const char *st2="Hello";

if(st1==st2){
    printf("True");
}
else{
    printf("False");
}
return 0;
}

Now in first code char array become const. In the first code I got False as optput. And in second code its true. Thank in advance

You compare pointers, not content. Second can also return false Btw. — Jarod42, Feb 05 '18 at 12:48
Second returns true is because this compile is clever to figure out these 2 strings are the same. — llllllllll, Feb 05 '18 at 12:50
@manni66 Don't agree. `stdio.h` is much more flexible in formatting. — llllllllll, Feb 05 '18 at 12:52
@liliscent yes, it's so flexible that you can easily shoot yourself into the foot. — , Feb 05 '18 at 12:56
@manni66: It would appear that way. It's ruined my answer on a technical point. — Bathsheba, Feb 05 '18 at 13:07

Bathsheba · Answer 1 · 2018-02-05T13:08:05.773

== does not compare the string contents.

In the first snippet st1 are st2 char[6] types with automatic storage duration, and you are allowed to modify the string contents. When using == these types decay to char*. Their addresses must be different, so == will yield false.

In the second snippet, the string literals are read only, in C they are still formally char[6] (cf. C++ where they are const char[6] types) although the behaviour on attempting to modify the contents is undefined. Using const char* types for them is perfectly acceptable and reasonable. Because the contents are read only, the compiler might use the same string and so st1 and st2 might point to the same location in memory. In your case, that is happening, and the result of == is true.

score -1 · Answer 2 · answered Feb 05 '18 at 12:51

-1

The second is true because you are using an optimizing compiler. Since str1 and str2 are pointers, it makes them point to the same string, thus saving a little memory.

answered Feb 05 '18 at 12:51

shawnhcorey

3,545
1
15
17

Nikita Smirnov · Accepted Answer · 2018-02-05T15:23:17.700

-2

Totally wrong answer for both C and C++. Pls, vote for delete.

edited Feb 05 '18 at 15:23

answered Feb 05 '18 at 12:55

Nikita Smirnov

813
8
17

3

This is not entirely accurate. In the first snippet, `st1` and `st2` are not pointers, they are arrays. In the second one, they are pointers to `char`, not pointers to arrays. And things they point to most probably are not on the heap. – HolyBlackCat Feb 05 '18 at 13:00
C does not have a stack. Please provide a reference to the standard stating otherwise. And HolyBlackCat is right, arrays are not pointers nor vice-vera. – too honest for this site Feb 05 '18 at 13:38
And @Olaf is rarely wrong (I've never known them to be) either. This answer is defective. – Bathsheba Feb 05 '18 at 15:03

Why first code output is false and true in second code?

3 Answers3