什么决定临时对象的生命周期何时扩展到 const 引用或右值引用?

What determines when the lifetimes of temporaries get extended into const references or rvalue references?

鉴于:

struct hurg { ... };

hurg get_hurg() { return hurg(); }
hurg&& get_mhurg() { return hurg(); }

我的理解 and experimenting 表明以下是 不是 未定义的行为(编辑:感谢答案,事实证明我错了 get_mhurg() 示例 未定义的行为):

{
    const hurg& a = get_hurg(); 
    hurg&& b = get_hurg();
    const hurg& c = get_mhurg(); 
    hurg&& d = get_mhurg();
    // do stuff with a, b, c, d
}
// a, b, c, d are now destructed

get_hurg()get_mhurg()返回的临时hurg对象的生命周期延长到作用域结束。

但是,在(来自 的函数)的情况下:

template <typename T>
auto id(T&& x) -> decltype(auto) { return decltype(x)(x); }    

像这样使用它:

{
    const hurg& x = id(hurg()); 
    // the hurg() 'x' refers to is already destructed

    hurg&& y = id(hurg());
    // the hurg() 'y' refers to is already destructed

    // undefined behavior: use 'x' and 'y'
}

在这种情况下,hurg 的生命周期不会 延长。

通常什么时候延长临时对象的生命周期?并且,特别是,什么时候将函数的结果绑定到 const 左值引用或右值引用是安全的?

更具体地说,id 案例中到底发生了什么?

来自[class.temporary]:

There are two contexts in which temporaries are destroyed at a different point than the end of the fullexpression. The first context is when a default constructor is called to initialize an element of an array [...]

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 of a sub-object to which the reference is bound persists for the lifetime of the reference except:
(5.1) — A temporary object bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.
(5.2) — The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not extended; the temporary is destroyed at the end of the full-expression in the return statement.
(5.3) — A temporary bound to a reference in a new-initializer (5.3.4) persists until the completion of the full-expression containing the new-initializer.

所以有两件事。首先, get_mhurg 是未定义的行为。您要返回的临时文件的生命周期 不会 延长。其次,传递给 id 的临时变量持续到包含函数调用的完整表达式结束, 但没有进一步 。与 get_mhurg 一样,临时文件不会被完全扩展。所以这也是未定义的行为。

My understanding and experimenting shows that the following is not undefined behavior:

嗯,你的理解和实验是错误的。明智的提示 - 不要尝试未定义的行为。例如,您的实验可能会随机碰巧表明它是明确定义的行为,这是未定义行为的可能结果之一。

使用 get_mhurg() 的 return 值是未定义的行为。您正在访问其创建范围之外的临时文件 hurg()

id 的情况表明延长生命周期是愚蠢的。

具体来说,生命周期延长仅直接适用于。它从不适用于任何类型的引用。您必须使用一个值来初始化引用。 id 没有 return 值,因此不应用生命周期延长。