Why Does a Virtual Function's Default Argument Use the Base Class's Value Instead of the Derived Class's?

Understanding Virtual Function Default Argument Behavior

In C , virtual functions allow derived classes to override the implementations of functions defined in base classes. However, the default argument behavior for virtual functions can be confusing.

Problem Description:

Consider the following code snippet:

class B {
public:
    B();
    virtual void print(int data=10) {
        cout << endl << "B--data=" << data;
    }
};

class D:public B {
public:
    D();
    void print(int data=20) {
        cout << endl << "D--data=" << data;
    }
};

int main() {
    B *bp = new D();
    bp->print();
    return 0;
}

Expected Output:

[ D--data=20 ]

Actual Output:

[ D--data=10 ]

Explanation:

According to the C standard (8.3.6.10), when calling a virtual function through a pointer or reference, the default arguments are derived from the static type of the pointer or reference, not the derived class's overriding function.

In this case, bp is a pointer of type B, so the default argument of B::print (which is 10) is used, overriding the default argument of D::print (which is 20).

Therefore, the output is D--data=10 instead of the expected D--data=20.

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