const member function clarification needed

Question

I'm a little confused as to why this code compiles and runs:

class A { private: int* b; public: A() : b((int*)0xffffffff) {} int* get_b() const {return this->b;} }; int main() { A a; int *b = a.get_b(); cout<<std::hex<<b<<endl; return 0; }

Output of running this code is FFFFFFFFas well... unexpected by me. Shouldn't this->b return const int* since it is in a const member function? and therefore the return line should generate a compiler-cast error for trying to cast const int* to int*

Obviously there's a gap here in my knowledge of what const member functions signify. I'd appreciate if someone could help me bridge that gap.

Bo Persson · Accepted Answer · 2011-04-27 15:03:14Z

4

No, the member is an int* const (as seen from the const function), which is totally different.

The pointer is const, not the object pointed to.

answered Apr 27, 2011 at 15:03

Bo Persson

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2 Comments

eladidan Over a year ago

Right. That explains it. So if b was int and get_b would try and return this->b then I would get the cast conversion error. Right?

Bo Persson Over a year ago

If b was an int and you tried to return &b, the compiler would complain on that, yes.

Xeo · Accepted Answer · 2011-04-27 15:02:00Z

The const part of a member function just says that the function is allowed to be called when the this pointer (aka the object on which it is called) is const. It got nothing to do with the return value.

class A{ public: void non_const_func(){} void const_func() const {} }; int main(){ A a; a.const_func(); // works a.non_const_func(); // works too const A c_a; c_a.const_func(); // works again c_a.non_const_func(); // EEEK! Error, object is const but function isn't!

}

user2100815 · Accepted Answer · 2011-04-27 15:02:37Z

The function returns an integer pointer by value - you can't change the class member it is a copy of via this value, so there is no const violation.

Robb · Accepted Answer · 2011-04-27 15:13:11Z

Putting const after the function declaration like you have tells the compiler "Hey, I promise not to modfy *this!". Your method is simply an accessor.

See C++ FAQ LITE 18.10

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