Question

在下面的程序中，是行

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;
}

Answer 1

这是允许的，但它没有意义，因为编译器会为你打电话。不过，我恐怕今天早上不想做标准拖网。

Answer 2

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（＆lt; =＆gt; -std = c ++ 98）

调用基类构造函数

2 个答案: