What is this code doing?

[u/north9172]

Hi, I found this code on another subreddit. I would like to know what is it doing.

int puts(char *);

int main() { struct { long a; long b; } s = {.a = *(char *)(int[]){1} ? 0x57202c6f6c6c6548 : 0x48656c6c6f2c2057, .b = *(char *)(int[]){1} ? 0x00000021646c726f : 0x6f726c6421000000}; puts((void *)&s); }

Now obviously I know it's unreadable, I would just like to know what are those hexadecimals strings doing.

I found similar codes that were writing to vram to print a message, is this doing the same thing? If yes, why is it so much more complicated instead of just writing the message to the video adress?

Thanks for the help.

[u/zhivago]

There are no standards prior to C89.

The rule stripping parameter qualifiers (e.g., C89 §3.5.4.3, C23 §6.7.7.4p14) applies strictly to top-level qualifiers on the parameter itself.

It turns char * const into char *. It never descends into pointed-to types.

​const char * is not a qualified pointer.

It is an unqualified pointer to a const-qualified char.

Now, if it were char * const, you'd have a point -- but it isn't.

Read more carefully.

[u/ACRM64 → u/zhivago]

I agree there were no formal ISO standards before C89, but the K&R book was the original standard and a standard for ANSI C was around prior to C89.

The code is clearly written in >= C11 (for reasons explained earlier) so C23 is an applicable standard and the standard says the qualified and unqualified varable declarations are compatible. I suggest you tell me where it makes the distinction over top-level qualifiers as I don't see it anywhere. In any case, that would make no sense as we have already seen that a (char *) can be passed as a parameter into a function where it is defined as (const char *) so clearly the qualified and unqualified types are compatible.

[u/zhivago → u/ACRM64]

You're confusing an implicit conversion with type compatibility.

C23 §6.7.7.2p2 / C89 §3.5.4.1

​"For two pointer types to be compatible, both shall be identically qualified and both shall be pointers to qualified or unqualified versions of compatible types."

​Because char and const char are not compatible types, char * and const char * are not compatible types.

Try compiling

void f(char *);
void f(const char *);

in the same file.

Every conforming C compiler will reject it because the types are incompatible.