与const和非成员的联合？ [英] An union with a const and a nonconst member?

问题描述

这似乎是未定义的行为

  union A {
 int const x; 
 float y; 
}; 
 
 A a = {0}; 
 a.y = 1; 
  

规格说

在具有静态，线程或自动存储持续时间的const对象占用的存储位置创建一个新对象，或者在这样的const对象在其生命周期结束之前占用的存储位置会导致未定义的行为。 / p>

但是没有编译器警告我，而这很容易诊断错误。

解决方案

最新的C ++ 0x草案标准明确说明了这一点：

在联合中，最多有一个非静态数据成员可以在任何
时间处于活动状态，也就是说，非静态数据成员可以随时将
存储在联合中。

>

  ay = 1; 
  

很好，因为它会更改活动成员 x 到 y 。如果您随后将 ax 作为右值引用，则该行为将未定义：

  cout < a.x<< endl; //未定义！ 
  

由于您没有创建任何新对象， p>

This appears to be undefined behavior

union A {
  int const x;
  float y;
};

A a = { 0 };
a.y = 1;

The spec says

Creating a new object at the storage location that a const object with static, thread, or automatic storage duration occupies or, at the storage location that such a const object used to occupy before its lifetime ended results in undefined behavior.

But no compiler warns me while it's an easy to diagnose mistake. Am I misinterpreting the wording?

解决方案

The latest C++0x draft standard is explicit about this:

In a union, at most one of the non-static data members can be active at any time, that is, the value of at most one of the non-static data members can be stored in a union at any time.

So your statement

a.y = 1;

is fine, because it changes the active member from x to y. If you subsequently referenced a.x as an rvalue, the behaviour would be undefined:

cout << a.x << endl ; // Undefined!

Your quote from the spec is not relevant here, because you are not creating any new object.

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