class A : public B
{
...
}
// case I : explicitly call the base class default constructor
A::A() : B()
{
...
}
// case II : don't call the base class default constructor
A::A() // : B()
{
...
}
案例II是否等于案例I?
对我而言,我认为在案例II中未调用基类B的默认构造函数。然而,尽管仍然持有这种看法,我进行了一项测试,证明了相反的结果:
class B
{
public:
B()
{
cout << "B constructor" << endl;
}
};
class A : public B
{
public:
A()
{
cout << "A constructor" << endl;
}
};
int _tmain(int argc, _TCHAR* argv[])
{
A a;
return 0;
}
// VS2008输出结果
B constructor
A constructor
Press any key to continue . . .