What does ((void (*)())buf)(); mean?
CFunctionPointersFunction PointersC Problem Overview
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?
C Solutions
Solution 1 - C
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.
Solution 2 - C
pointer buf is converted to the pointer to void function taking unspecified number of parameters and then dereferenced (ie function called).
Solution 3 - C
It's a typecast, followed by a function call. Firstly, buf is cast to the pointer to a function that returns void. The last pair of parenthesis means that the function is then called.
Solution 4 - C
It casts the character array to a pointer to a function taking no arguments and returning void, and then calls it. Dereferencing the pointer is not required due to how function pointers work.
An explanation:
That "character array" is actually an array of machine code. When you cast the array to a void (*)() and call it, it runs the machine code inside of the array. If you provided the array's contents I could disassemble it for you and tell you what it's doing.