Calling function from its memory address?

[Justin H]

I searched the internet for a method of doing this, and came across this neat little trick:

typedef void func(void);
func* f = (func*)0xdeadbeef;
f();

And I'm a little stuck here on understanding the concept of this. Is the 0xdeadbeef suppose to act as a null/garbage value?

Also, can

func* f = (func*)0xdeadbeef;

just be written as:

func *f = (func*)0xdeadbeef;

Also, what's the point of (func*)?

Sorry for the rather newbie question, however I can't quite understand what's happening here. Thanks for the help!

[whiteflags]

well normally you would do something like:

void actual_func(void);
typedef void ( *func)(void);

func fp = &actual_func;

// Call func.
fp();

I don't know why you would want to refer to a specific address instead of a specific function name instead. I mean, especially with modern systems, how are you to know the real address? I suppose that is part of the rationale for why C++ doesn't let you assign addresses like numbers, you have to cast them to a pointer type. Function pointers themselves are still useful on some occasions; this may or may not be one of them.

[Justin H]

Ah, thank you. I guess I just got a little confused on what I was wanting to do there.

Your example is actually what I wanted to do. Thanks.

[blakeo_x]

0xdeadbeef is just a placeholder address for example purposes, similar to "foo", "bar", etc. The address has no significant meaning in C++.

The location of the pointer operator can indeed be in either of the configurations you gave, or even in the form "func * f" with a space on both sides.

The point of the "(func*)" is to typecast the address to be of the func* type (effectively void because of the preceeding typedef, which is a special type of pointer). See Pointers - C++ Documentation for more on function pointers and void pointers.

Edit: I do believe whiteflag's suggestion is a more proper usage, but unless I'm mistaken, both should "work".