Calling a const function rather than its non-const version

Question

I tried to wrap something similar to Qt\'s shared data pointers for my purposes, and upon testing I found out that when the const function should be called, its non-const ve

Answer 1

If you have two overloads that differ only in their const-ness, then the compiler resolves the call based on whether *this is const or not. In your example code, test is not const, so the non-const overload is called.

If you did this:

testType test;
const testType &test2 = test;
test2->x();

you should see that the other overload gets called, because test2 is const.