const 引用类成员是否会延长临时对象的寿命?
为什么会这样:
#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;
}
给出输出:
答案是:
代替:
答案是:四个
推荐答案
只有 local const 引用才能延长寿命.
Only local const references prolong the lifespan.
标准在第 8.5.3/5 节 [dcl.init.ref] 中关于引用声明的初始化程序部分指定了此类行为.您示例中的引用绑定到构造函数的参数 n，并且当对象 n 绑定到超出范围时变为无效.
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.
生命周期延长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]:
第二个上下文是引用绑定到临时的.引用绑定到的临时对象或作为临时对象绑定的子对象的完整对象的临时对象将在引用的生命周期内持续存在，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会一直存在，直到构造函数退出.在函数调用(第 5.2.2 节 [expr.call])中临时绑定到引用参数会一直存在，直到包含调用的完整表达式完成为止.
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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