问题描述

在问这个问题时，我学会了const引用到临时对象在 C++ 中是有效的:

While asking this question, I learned const reference to a temporary object is valid in C++:

int main ()
{
  int a = 21;
  int b = 21;

  //error: invalid initialization of non-const reference
  //int     & sum = a + b;e [...]

  //OK
  int const & sum = a + b;

  return sum;
}

但是在下面的例子中，const 引用 refnop 引用了一个被破坏的临时对象.我想知道为什么?

But in the following example, the const reference refnop refers to a destroyed temporary object. I wonder why?

#include <string>
#include <map>

struct A
{
   // data 
   std::map <std::string, std::string>  m;
   // functions
   const A& nothing()           const { return *this;    }
   void init()                        { m["aa"] = "bb";  }
   bool operator!= (A const& a) const { return a.m != m; }
};

int main()
{
  A a;
  a.init();

  A const& ref    = A(a);
  A const& refnop = A(a).nothing();

  int ret = 0;
  if (a != ref)     ret += 2;
  if (a != refnop)  ret += 4;

  return ret;
}

使用 GCC 4.1.2 和 MSVC 2010 测试，返回 4；

Tested using GCC 4.1.2 and MSVC 2010, it returns 4;

$> g++ -g refnop.cpp
$> ./a.out ; echo $?
4

ref 和 refnop 之间的区别在于对 nothing() 的调用实际上什么都不做.看来这次调用之后，临时对象就被销毁了！

The difference between ref and refnop is the call to nothing() which does really nothing. It seems after this call, the temporary object is destroyed!

我的问题:
为什么在refnop的情况下，临时对象的生命周期与其const引用不同?

My question:
Why in the case of refnop, the life time of the temporary object is not the same as its const reference?

推荐答案

当临时对象绑定到第一个引用时，临时对象的生命周期扩展只能执行一次.之后，引用指向临时对象的知识就消失了，因此无法再延长生命周期.

The lifetime-extension of a temporary object can be performed only once, when the temporary object gets bound to the first reference. After that, the knowledge that the reference refers to a temporary object is gone, so further lifetime extensions are not possible.

让你困惑的案例

A const& refnop = A(a).nothing();

类似于这种情况:

A const& foo(A const& bar)
{
    return bar;
}
//...
A const& broken = foo(A());

在这两种情况下，临时变量都绑定到函数参数(nothing() 的隐式 this， 的 barfoo()) 并将其生命周期扩展"到函数参数的生命周期.我将'extended'放在引号中，因为临时文件的自然生命周期已经更长，因此不会发生实际的扩展.

In both cases, the temporary gets bound to the function argument (the implicit this for nothing(), bar for foo()) and gets its lifetime 'extended' to the lifetime of the function argument. I put 'extended' in quotes, because the natural lifetime of the temporary is already longer, so no actual extension takes place.

因为生命周期扩展属性是不可传递的，返回一个引用(恰好指向一个临时对象)不会进一步延长临时对象的生命周期，结果是 refnop和 broken 最终指向不再存在的对象.

Because the lifetime extension property is non-transitive, returning a reference (that happens to refer to a temporary object) will not further extend the lifetime of the temporary object, with as result that both refnop and broken end up referring to objects that no longer exist.

这篇关于在函数作用域(生命周期)之后，对临时对象的 const 引用被破坏的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！