Question

class ClassB {
    int m_b;
public:
    ClassB(int b) : m_b(b) {}
    void PrintClassB() const {
        cout << "m_b: " << m_b << endl;
    }
};


int main(int argc, char* argv[])
{
    const ClassB &af = ClassB(1);
    af.PrintClassB(); // print m_b: 1 with vs2008 & gcc 4.4.3
}

Given the above code, I have difficulties to understand this snippet:

Q1> What does this line mean?

const ClassB &af = ClassB(1);

Here is my understanding:

af refers to a temporary variable ClassB(1) and after the

execution of this line, the temporary variable is destroyed and af
refers to an undefined variable. During this procedure, no
copy-constructor is called.

Then why we can still issue the following statement and obtain the results?

af.PrintClassB(); // print m_b: 1 with vs2008 & gcc 4.4.3

Answer 1

const ClassB &af = ClassB(1);

const here extend the life time of the temporary object (i.e., ClassB(1)) being created. It’s scope lasts until af falls out of scope;

af.PrintClassB(); // print m_b: 1 with vs2008 & gcc 4.4.3

This is because, af is nothing but the temporary object’s reference which was constructed passing 1 to it’s constructor.