Nonstatic member as a default argument of a nonstatic member function

Question

struct X
{
   X():mem(42){}
   void f(int param = mem) //ERROR
   {
      //do something
   }
private: 
   int mem;
};

Can anyone give me just one

Answer 1

For one reason, because f is public, but mem is private. As such, code like this:

int main() { 
    X x;
    x.f();
    return 0;
}

...would involve outside code retrieving X's private data.

Aside from that, it would (or at least could) also make code generation a bit tricky. Normally, if the compiler is going to use a default argument, it gets the value it's going to pass as part of the function declaration. Generating code to pass that value as a parameter is trivial. When you might be passing a member of an object (possibly nested arbitrarily deeply) and then add in things like the possibility of it being a dependent name in a template, that might (for example) name another object with a conversion to the correct target type, and you have a recipe for making code generation pretty difficult. I don't know for sure, but I suspect somebody thought about things like that, and decided it was better to stay conservative, and possibly open thins up later, if a good reason was found to do so. Given the number of times I've seen problems arise from it, I'd guess it'll stay the way it is for a long time, simply because it rarely causes problems.