Why does calling a member function as if it were static (when it's not) work in C++?

Question

According to my understanding of C++, the following code is erroneous.

#include <iostream>

class Test {
private:
  int num_;

public:
  Test(int n) : num_(n) {}
  void printNum() { std::cout << num_ << '\n'; }
  void weird() { Test::printNum(); }
};

int main() {
  Test t(10);
  t.weird();
  return 0;
}

Test::weird() calls Test::printNum() as it would a static member function. However, Test::printNum() accesses an instance attribute and is clearly not static. Yet, the code compiles and runs to output 10.

My compiler is Apple LLVM version 7.0.0 (clang-700.1.76)

What am I missing?

Answer 1

What am I missing?

You're wrong about this:

Test::weird() calls Test::printNum() as a class (static) method.

It's not being called as a static method. It can't be, because it's not one.

Within a member function, you do not need an object reference or pointer (e.g. this) to call another member function. So you can write either of these:

this->printNum();
printNum();

The full name of printNum is actually Test::printNum, so you can do these too:

this->Test::printNum();
Test::printNum();

Outside of a member function, the second option in both cases is wrong because the function is not static and you did not provide an object reference or pointer.

Outside of a member function you'd also have to write the Test:: otherwise the compiler wouldn't know which printNum you're talking about, but that in itself doesn't force the call to be a "static" call. It'll be a static call iff the member function is static; period!