执行void（char *）函数作为void（void）

Question

void print( char * str ){ printf("%s",str); }

void some_function(){
    ... Loads a file into `char * buffer` ...
    ... Stores the function print before `buffer` address ...
    ((void(*)(void))buffer();
}

In the file, there is a "Hello World" in it, and some unreadable characters. Executing the buffer will print "Hello World".

I know you can execute a pointer like this:

void (*foo)(int) = &bar; // such that void bar(int)
(*foo)(123);

But executing void(int) function as a void(void) function with its function and parameters in memory is new to me.

Is there a standard of how a function look like in memory (like a string is terminated by a null character) such that you can execute it in this way?

Answer 1

But executing void(int) function as a void(void) function with its function and parameters in memory is new to me.

Good. It's undefined behavior and you should never do it. It might happen to work on some implementation, but there's no guaranteed result.

Is there a standard of how a function look like in memory (like a string is terminated by a null character) such that you can execute it in this way?

No, there isn't. More relevantly, there is no standard for how arguments are passed to functions -- it's up to the implementation, and there is no requirement that it be implemented in such a way that functions with different types are compatible.

But in this case, the programmer is taking advantage of knowledge of what a function looks like in memory, and has prepared the file in such a way that it does the right thing when loaded into memory and run in this particular implementation. This is, after all, what loaders do ... they read programs from files into memory and then execute them. This method is also used by exploits that take advantage of bugs in programs to load nefarious code into the program's memory and get the program to execute it.