在初始化列表中初始化引用 [英] initialize reference in initialization list

问题描述

我被告知引用变量必须在初始化列表中初始化，但为什么这是错误的？

  class Foo 
 {
 public：
 Foo（）：x（0）{
y = 1; 
} 
 private：
 int& X; 
 int y 
}; 
  

因为0是临时对象？如果是这样，什么样的对象可以引用绑定？

解决方案

0不是一个左值，它是一个右值。

如果你引用 const ，那么你可以修改它。 code>，它会按预期工作。考虑这个：

  int& x = 0; 
 x = 1; // wtf :( 
  

这显然是一个没有，但 const& ; 可以绑定到临时（rvalues）：

  const int& x = 0; 
x = 1; // protected :) [不会编译] 
  

临时的生命期在构造器完成时结束。如果您为自己的常数使用静态存储，您将是安全的：

  class Foo 
 {
 public：
 static const int Zero = 0; 
 
 Foo（）：x（Zero）// Zero有存储
 {
 y = 1; 
} 
 private：
 const int& X; 
 int y 
}; 
  




I was told the reference variable must be initialized in the initialization list, but why this is wrong?

   class Foo
    {
    public: 
        Foo():x(0) {      
         y = 1;
        }
    private:
        int& x;
        int y;
    };

Because 0 is a temporary object? If so, what kind of object can reference be bound? The object which can take an address?

Thanks!

解决方案

0 is not an lvalue, it's an rvalue. You cannot modify it, but you're trying to bind to a reference where it could be modified.

If you make your reference const, it will work as expected. Consider this:

int& x = 0;
x = 1; // wtf :(

This obviously is a no-go. But const&'s can be bound to temporaries (rvalues):

const int& x = 0;
x = 1; // protected :) [won't compile]

Note that the life-time of the temporary is ended at the completion of the constructor. If you make static-storage for your constant, you'll be safe:

class Foo
{
public:
    static const int Zero = 0;

    Foo() : x(Zero) // Zero has storage
    {
        y = 1;
    }
private:
    const int& x;
    int y;
};

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