CODEWITHSUNDEEP
c++pointers
bcf
bcf

Reputation: 2134

What are Function Pointers Ponting To?

Consider the following code and output:

#include <iostream>

int Add(int a, int b) {
  return a + b;
}

int Subtract(int a, int b) {
  return a - b;
}

int main() {
  int (*fn1)(int, int);
  int (*fn2)(int, int);
  fn1 = &Add;
  fn2 = &Subtract;

  std::cout << "fn1 = " << fn1 << "\n";
  std::cout << "*fn1 = " << *fn1 << "\n";
  std::cout << "fn2 = " << fn2 << "\n";
  std::cout << "*fn2 = " << *fn2 << "\n";
}

Output:

fn1 = 1
*fn1 = 1
fn2 = 1
*fn2 = 1

As a pointer, I would expect fn1 and fn2 to be memory addresses, and I don't really know what to expect for *fn1 and *fn2. According to Wikipedia, "a function pointer points to executable code within memory", but I don't see how "executable code within memory" corresponds with "1".

Upvotes: 0

Views: 71

Answers (2)

Employed Russian
Employed Russian

Reputation: 213955

What are Function Pointers Ponting To?

It depends on what kind of function pointer it is. For non-member function pointer (as in your example), the pointer usually points to the actual code for the function (though on some architectures, such as PowerPC, it may point into a special "function descriptor" instead):

To examine that value, you should print it as void *, and not as bool (which is what you have done in your example).

std::cout << "fn1 = " << (void*)fn1 << std::endl;

Another way to examine it is using a debugger. For example, in GDB:

(gdb) p/a fn1
(gdb) p/a fn2

Upvotes: 3

Kerrek SB
Kerrek SB

Reputation: 477640

*fn1 is a function lvalue, and such a value decays to a pointer to that function in most circumstances, including when passed as a function call argument.

You can dereference the result of the decay to get another function lvalue, ad infinitum: *****fn1, ***********fn1, etc.

Upvotes: 3

Related Questions