What's the right behavior when auto meets polymorphism and virtual function?

Question

class B {
    public:
    virtual void f(){
        printf("B\n");
    }
};
class D : public B { 
    public:
    void f() {
        printf("D\n");
    }
};

int main(void)  
{  
    B* d = new D();
    d->f();
    auto b = *d; 
    b.f();
}

for d->f();, the output is D. this is right. But for b.f();, the output is B. Is this right?

It looks like you've forgotten to _override_ your virtual function. — Disillusioned, Jul 12 '16 at 05:43
`auto` type deduction and for that matter template type deduction also works on compile time /static type not on its runtime type. — Arunmu, Jul 12 '16 at 05:45

score 14 · Accepted Answer · edited May 23 '17 at 10:28

14

Is this right?

It's right, the type is deduced at compile time. auto uses the same rules of template argument deduction for type deduction, based on the static type, dynamic polymorphism won't be considered.

For this case, the type of d is B*, then type of *d is B, so the type of b is just B. Then *d will be slicing copied to b, for b.f() B::f() should be invoked.

The code is equivalent to the following which might be clearer.

B b = *d;
b.f();

edited May 23 '17 at 10:28

Community

1
1

answered Jul 12 '16 at 05:37

songyuanyao

169,198
16
310
405

However, please note that if it were `auto & b = *d;` (or `auto && b = *d;` ) , then `b.f()` would have invoked the overridden f() in the derived class. – Nawaz Jul 12 '16 at 06:36

What's the right behavior when auto meets polymorphism and virtual function?

1 Answers1