What does ((void (*)())buf)(); mean?

c function pointers function-pointers
sh.3.ll picture sh.3.ll · Jan 14, 2020 · Viewed 7.9k times · Source

I am solving a binary exploitation challenge on picoCTF and came across this piece of code:

((void (*)())buf)();

where buf is a character array.

I solved the challenge but can't seem to understand what exactly it's doing. I looked at this thread but I couldn't make it out.

What does ((void (*)())buf)(); mean?

Answer

Some programmer dude picture Some programmer dude · Jan 14, 2020

void (*)() is a type, the type being "pointer to function that takes indeterminate arguments and returns no value".

(void (*)()) is a type-cast to the above type.

(void (*)())buf casts buf to the above type.

((void (*)())buf)() calls the function (passing no arguments).

In short: It tells the compiler to treat buf as a pointer to a function, and to call that function.

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