const_cast rules in c++

Question

struct foo
{
    const int A;
    int B;
    foo() : A(10), B(20) {}
};

void main()
{
    foo f1;
    const_cast<int&>(f1.A) = 4; //line 1
    const foo f2;
    const_cast<int&>(f2.B) = 4; //line 2
}

Do both line 1 and 2 exhibit undefined behaviour? Would the behavior be different if f1 and f2 were shared_ptr of types listed in code above?

Boutte

It does not matter what data type is loosing constness, problem is compiler can assume that object does not change and produce relevant code. – Undertrump

Answer 1

The behaviour both const_cast<int&>(f1.A) = 4 and const_cast<int&>(f2.B) = 4 are undefined.

If an object is originally defined as const, and you cast away that const-ness and attempt to modify the object, the behaviour is undefined.

Saccharase

@YSC: Hum. Could've sworn f2 was non-const originally. Still, only a little edit needed. Thanks. – Saccharase

Edits making non-const object now const (even sneaky edits made on Fridays) are welcome. I wish there was more of them at work. – Offenseless

As an addendum, if an object is originally non-const, and then passed through a const pointer or const reference, and then const_cast back to non-const, I believe that is well-defined behavior. But, in my opinion, it is still a bad code smell to const_cast away const-ness even if the developer KNOWS it is a non-const object really really. – Blackboard

@Eljay: Yes you are correct, both factually and stylistically. It's only if you attempt to modify the object do you get undefined results. – Saccharase

@Blackboard there's the school of thought for de-duplicating getter's of the style class Foo { const Bar& do_get() const { /* single impl */ } public: Bar& get() { return const_cast<Bar&>(do_get()); } const Bar& get() const { return do_get(); } }; – Acid