返回临时对象并绑定到 const 引用 [英] Returning temporary object and binding to const reference

问题描述

可能的重复:
const 引用是否会延长临时引用的寿命?

我的编译器不会抱怨将临时分配给 const 引用:

My compiler doesn't complain about assigning temporary to const reference:

string foo() {
  return string("123");
};

int main() {
  const string& val = foo();
  printf("%s
", val.c_str());
  return 0;
}

为什么?我认为从 foo 返回的字符串是临时的，val 可以指向生命周期已经结束的对象.C++ 标准是否允许这样做并延长返回对象的生命周期?

Why? I thought that string returned from foo is temporary and val can point to object which lifetime has finished. Does C++ standard allow this and prolongs the lifetime of returned object?

推荐答案

这是一个 C++ 特性.该代码是有效的，并且完全符合它的功能.

This is a C++ feature. The code is valid and does exactly what it appears to do.

通常，临时对象仅持续到它出现的完整表达式的结尾.但是，C++ 特意指定将临时对象绑定到堆栈上对 const 的引用会将临时对象的生命周期延长至引用本身的生命周期，从而避免常见的悬空引用错误.在上面的例子中， foo() 返回的临时值一直存在到右花括号为止.

Normally, a temporary object lasts only until the end of the full expression in which it appears. However, C++ deliberately specifies that binding a temporary object to a reference to const on the stack lengthens the lifetime of the temporary to the lifetime of the reference itself, and thus avoids what would otherwise be a common dangling-reference error. In the example above, the temporary returned by foo() lives until the closing curly brace.

P.S:这仅适用于基于堆栈的引用.它不适用于作为对象成员的引用.

P.S: This only applies to stack-based references. It doesn’t work for references that are members of objects.

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