Static_cast and virtual methods in c++

Question

in the following code, due to name() being virtual, I would expect that the method of derived struct will be called. Conversely, whats get written out is "A". Why?

#include <iostream>
using namespace std;
struct A {
    virtual string name() { return "A"; }
};
struct B : A {
    string name() { return "B"; }
};
int main (int argc, char *argv[]) {
    B b;
    cout << static_cast<A>(b).name() << endl;
    return 0;
}

Polymorphism only applies to method calls via pointers or references. You've essentially done `A tmp = b; cout << tmp.name() << endl;`. — Oliver Charlesworth, Jun 20 '15 at 21:03

score 6 · Accepted Answer · answered Jun 20 '15 at 21:09

6

static_cast<A>(b) creates a temporary variable of type A constructed from b. So calling name() indeed invokes A::name().

In order to observe polymorphic behavior you may do

static_cast<A&>(b).name()

answered Jun 20 '15 at 21:09

Anton Savin

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score 1 · Answer 2 · edited May 23 '17 at 12:14

1

What you are experiencing is called slicing. Essentially the static cast will cut the B part away from the object and you keep just an A.

What is object slicing?

edited May 23 '17 at 12:14

Community

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answered Jun 20 '15 at 21:11

Christopher Oezbek

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Steephen · Answer 3 · 2015-06-20T21:10:55.877

0

As oliver Charlesworth mentioned in comment, you need pointer or refernce of the object required to see the impact of polymorphism. Then system will identify the dynamic type of the object and invoke the corresponding function. How you can invoke polymorphism using reference is explained in Anton Savin's answer.

For the expected result if you want to use pointer, you have to do as follows in main:

A *a= new B();
cout << a->name() << endl;

edited Jun 20 '15 at 21:10

answered Jun 20 '15 at 21:08

Steephen

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Why complicate things with dynamic allocation? – Ben Voigt Jun 20 '15 at 21:09

Static_cast and virtual methods in c++

3 Answers3