Calling a const function rather than its non-const version

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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.

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