Why does this program hang?

Question

I have the following code which is seems to be lead to the infinite loop:

struct X
{
  void my_func( int ) { std::cout << "Converted to int" << std::endl; }
};

struct X2 : X
{
  void my_func( char value ) { my_func(value); }
};

What is the problem with it?

Answer 1

void my_func( char value ) { my_func(value); }

You're passing the value which is char so it resolves to calling the same method with accepts a char parameter. It becomes an endless loop.

Answer 2

void my_func( char value ) { my_func(value); }

right there, you've written a recursive function with no base case. I don't know too much about C++, but you need to somehow specify that you want to call X's my_func, not X2's(I'm assuming that's what you want to do.)

edit: To fix it, you need to cast value to an int

Answer 3

The program gets in an infinite loop. my_func() calls itself and there's no condition to exit out of it.

Answer 4

Your call my_func(value) is recursive. Did you mean super::my_func(value) ?

Answer 5

The issue occurs due to hiding. http://www.parashift.com/c++-faq-lite/strange-inheritance.html#faq-23.9

Answer 6

You need to explicitly call the base class' function, i.e.:

struct X
{
  void my_func( int ) { std::cout << "Converted to int" << std::endl; }
};

struct X2 : X
{
  void my_func( char value ) { X:my_func(value); }
};

By default, the compiler uses functions within the same class if present, as there is no way for it to know which one you actually want to use. By specifying BaseClass::Function within a derived class' method, the compiler will explicitly create a call to that base class's method, even if you have overridden.