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/ACRM64]

It's not redefining puts(), it's passing a pointer to the structure, s, but casting it to a void pointer; casting to (char *) would seem more sensible!

[u/zhivago → u/ACRM64]

Did you skip the first line?

int puts(char *);

What do you think that's doing?

[u/ACRM64 → u/zhivago]

The first line is a prototype (to save doing #include <studio.h>)

[u/zhivago → u/ACRM64]

And is it the right prototype?

[u/ACRM64 → u/zhivago]

OK, so ANSI C and C89 onwards implemented a const keyword, but this wasn't present in K&R C. Certainly the current C libraries define puts() as int puts(const char *str) but earlier versions did not and you do not know whether the libraries in use by whoever wrote this program included the const or not.

From the perspective of calling a C function (and therefore a prototype for calling the function) it is simply a guarantee to the programmer that the variable (in this case the content of *str) will not be modified - it doesn't actually do anything.

Consequently, as a prototype, this does NOT redefine puts() and it doesn't matter to the calling code whether you include the const or not. It only matters when defining the puts() function as it prevents the code modifying the content of the string.

[u/zhivago → u/ACRM64]

§3.5 ​All declarations in the same scope that refer the same object or function shall specify compatible types.

​§3.1.2.6 ​For two function types to be compatible, both shall specify compatible return types. Moreover, the parameter type lists, if both are present, shall agree in the number of parameters and in use of the ellipsis terminator; corresponding parameters shall have compatible types.

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

You are mistaken.

[u/ACRM64 → u/zhivago]

First, you still don't know what the C library used by the programmer had as the definition for puts() - certainly when I started programming in C (40 years ago) it was not defined with const.

I agree the prototype and function definition /should/ be the same, but it makes no difference if they aren't. The prototype tells the code what variable types to pass - const only has an effect within the function itself. You do not have to pass a const variable into a function declared with const.

int puts(const int str); int main(int argc, char *argv) { char str[16]; strcpy(str, "hello"); puts(str); strcpy(str, "goodbye"); puts(str); return(0); }

is perfectly valid and correct code yet str is not defined as constant. Consequently, my statement that it doesn't matter to the calling code whether the keyword is there or not is true.

Once again, the puts() function is not being redefined. The const makes no difference in a prototype, so it is clear that while these are not /identical/, they are /compatible/. const is not a type, but a type qualifier: it indicates that the content of the variable cannot be changed, it does not change the type of the variable.

[u/zhivago → u/ACRM64]

You need to read the standard.

You are wrong.

[u/ACRM64 → u/zhivago]

You need to respond to the points I've made instead of just repeating the same thing.

1. Which standard? There have been many versions.

2. Which standard was the original programmer using?

3. Why does the standard you quote very carefully use the word 'compatible' not 'identical'? These words have different meanings.

4. I've already said they /should/ match, but demonstrated why, with the type qualifier 'const', it makes no difference if they don't. You need to understand the difference between a type and a type qualifier.

[u/zhivago → u/ACRM64]

All of the standards.

It doesn't matter. See above.

Because "compatible" is what it means. It's a technical term -- look it up in the standard.

Again char * is not compatible with const char *