调用基类构造函数
在下面的程序中，该行是否
Derived(double y): Base(), y_(y)
正确/允许？也就是说，它遵循 ANSI 规则吗？
#include <iostream>
class Base
{
public:
Base(): x_(0)
{
std::cout << "Base default constructor called" << std::endl;
}
Base(int x): x_(x)
{
std::cout << "Base constructor called with x = " << x << std::endl;
}
void display() const
{
std::cout << x_ << std::endl;
}
protected:
int x_;
};
class Derived: public Base
{
public:
Derived(): Base(1), y_(1.2)
{
std::cout << "Derived default constructor called" << std::endl;
}
Derived(double y): Base(), y_(y)
{
std::cout << "Derived constructor called with y = " << y << std::endl;
}
void display() const
{
std::cout << Base::x_ << ", " << y_ << std::endl;
}
private:
double y_;
};
int main()
{
Base b1;
b1.display();
Derived d1;
d1.display();
std::cout << std::endl;
Base b2(-9);
b2.display();
Derived d2(-8.7);
d2.display();
return 0;
}
In the program below, is the line
Derived(double y): Base(), y_(y)
correct/allowed? That is, does it follow ANSI rules?
#include <iostream>
class Base
{
public:
Base(): x_(0)
{
std::cout << "Base default constructor called" << std::endl;
}
Base(int x): x_(x)
{
std::cout << "Base constructor called with x = " << x << std::endl;
}
void display() const
{
std::cout << x_ << std::endl;
}
protected:
int x_;
};
class Derived: public Base
{
public:
Derived(): Base(1), y_(1.2)
{
std::cout << "Derived default constructor called" << std::endl;
}
Derived(double y): Base(), y_(y)
{
std::cout << "Derived constructor called with y = " << y << std::endl;
}
void display() const
{
std::cout << Base::x_ << ", " << y_ << std::endl;
}
private:
double y_;
};
int main()
{
Base b1;
b1.display();
Derived d1;
d1.display();
std::cout << std::endl;
Base b2(-9);
b2.display();
Derived d2(-8.7);
d2.display();
return 0;
}
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评论（2）
这是允许的，但毫无意义，因为编译器会为您进行调用。不过，恐怕今天早上我不想进行标准拖网捕鱼。
It's allowed, but it's pointless, as the compiler will make the call for you. I'm afraid I don't feel like doing a Standard trawl this morning, though.
这是正确的，但不需要调用基类默认构造函数。
假设您使用的是 g++，您可能需要使用以下标志：-ansi (<=> -std=c++98)
This is correct but calls to the base class default constructors are not necessary.
Assuming you are using g++, you may want to use the following flag: -ansi (<=> -std=c++98)