const引用是否延长了临时的生命？ [英] Does a const reference prolong the life of a temporary?

问题描述

为什么：

#include <string>
#include <iostream>
using namespace std;

class Sandbox
{
public:
    Sandbox(const string& n) : member(n) {}
    const string& member;
};

int main()
{
    Sandbox sandbox(string("four"));
    cout << "The answer is: " << sandbox.member << endl;
    return 0;
}

输出：

答案是：

The answer is:

而不是：

答案是：四个

The answer is: four

推荐答案

只有 local const 可延长寿命。

8.5.3 / 5，[dcl.init.ref]，关于参考声明初始化器的部分。在你的例子中的引用绑定到构造函数的参数 n ，并且当对象 n 超出范围。

The standard specifies such behavior in §8.5.3/5, [dcl.init.ref], the section on initializers of reference declarations. The reference in your example is bound to the constructor's argument n, and becomes invalid when the object n is bound to goes out of scope.

生存期扩展不能通过函数参数传递。 §12.2/ 5 [class.temporary]：

The lifetime extension is not transitive through a function argument. §12.2/5 [class.temporary]:

第二个上下文是当引用绑定到临时时。引用绑定到的临时或临时，即临时绑定的子对象的完整对象将在引用的生存期内保持，除非下面指定。在构造函数的ctor-initializer（§12.6.2[class.base.init]）中临时绑定到引用成员，直到构造函数退出。在函数调用中的引用参数的临时绑定（§5.2.2[expr.call]）一直持续到包含调用的完整表达式完成。

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object to a subobject of which the temporary is bound persists for the lifetime of the reference except as specified below. A temporary bound to a reference member in a constructor’s ctor-initializer (§12.6.2 [class.base.init]) persists until the constructor exits. A temporary bound to a reference parameter in a function call (§5.2.2 [expr.call]) persists until the completion of the full expression containing the call.

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