Print address of virtual member function

问题:

I am trying to print the address of a virtual member function. If I know which class implements the function I can write:

print("address: %p", &A::func);

But I want to do something like this:

A *b = new B();

printf("address: %p", &b->func); 
printf("address: %p", &b->A::func);

However this does not compile. Is it possible to do something like this, perhaps looking up the address in the vtable at runtime?

回答1:

Currently there is no standard way of doing this in C++ although the information must be available somewhere. Otherwise, how could the program call the function? However, GCC provides an extension that allows us to retrieve the address of a virtual function:

void (A::*mfp)() = &A::func;
printf("address: %p", (void*)(b->*mfp));

...assuming the member function has the prototype void func(). This can be pretty useful when you want to cache the address of a virtual function or use it in generated code. GCC will warn you about this construct unless you specify -Wno-pmf-conversions. It's unlikely that it works with any other compiler.

回答2:

Pointers to member functions are not always simple memory addresses. See the table in this article showing the sizes of member function pointers on different compilers - some go up to 20 bytes.

As the article outlines a member function pointer is actually a blob of implementation-defined data to help resolve a call through the pointer. You can store and call them OK, but if you want to print them, what do you print? Best to treat it as a sequence of bytes and get its length via sizeof.

回答3:

Doesn't make a lot a of sense to me. If you have a normal function:

void f( int n ) {
}

then you can take its address:

f

but you cannot take the address of a function call, which is what you seem to want to do.

回答4:

From what I can tell in the standard, the only time you get dynamic binding is during a virtual function call. And once you've called a function, you're executing the statements within the function (i.e., you can't "stop halfway" into the call and get the address.)

I think it's impossible.