做右值引用允许悬挂引用？ [英] Do rvalue references allow dangling references?

问题描述

请考虑以下内容。

  #include< string& 
 using std :: string; 
 
 string middle_name（）{
 returnJaan; 
} 
 
 int main（）
 {
 string&& danger = middle_name（）; //？！ 
 return 0; 
} 
  

这不计算任何东西，但它编译没有错误，我发现混淆： danger 是一个悬挂的引用，不是吗？

解决方案

如果您的意思是可以创建悬挂，您可以创建悬挂引用？ 右值引用，那么答案是肯定的。

  string middle_name（）{
 returnJaan; 
} 
 
 int main（）
 {
 string&& nodanger = middle_name（）; // 好。 
 //临时的生命周期扩展为
 //到引用的生命周期。 
 return 0; 
} 
  

此处适用相同的规则，使此示例< a>（Herb Sutter的文章）安全。如果您使用 右值初始化引用，则tempoary对象的生命周期将延长到引用的生命周期。你仍然可以产生悬挂引用。例如，这不再安全：

  int main（）
 {
 string&& danger = std :: move（middle_name（））; // dangling reference！ 
 return 0; 
} 
  

因为 std :: move 返回 string&&& （ pure 右值）延长临时生命周期的规则不适用。这里， std :: move 返回所谓的 xvalue 。 xvalue 只是一个未命名的右值引用。因此，它可以引用任何东西，基本上不可能猜到返回的引用指向什么，而不查看函数的实现。

Consider the below.

#include <string>
using std::string;

string middle_name () {
    return "Jaan";
}

int main ()
{
    string&& danger = middle_name();   // ?!
    return 0;
}

This doesn't compute anything, but it compiles without error and demonstrates something that I find confusing: danger is a dangling reference, isn't it?

解决方案

Do rvalue references allow dangling references?

If you meant "Is it possible to create dangling rvalue references" then the answer is yes. Your example, however,

string middle_name () {
    return "Jaan";
}

int main()
{
    string&& nodanger = middle_name();   // OK.
    // The life-time of the temporary is extended
    // to the life-time of the reference.
    return 0;
}

is perfectly fine. The same rule applies here that makes this example (article by Herb Sutter) safe as well. If you initialize a reference with a pure rvalue, the life-time of the tempoary object gets extended to the life-time of the reference. You can still produce dangling references, though. For example, this is not safe anymore:

int main()
{
    string&& danger = std::move(middle_name());  // dangling reference !
    return 0;
}

Because std::move returns a string&& (which is not a pure rvalue) the rule that extends the temporary's life-time doesn't apply. Here, std::move returns a so-called xvalue. An xvalue is just an unnamed rvalue reference. As such it could refer to anything and it is basically impossible to guess what a returned reference refers to without looking at the function's implementation.

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