c++ - 临时右值引用对象的生命周期

Question

假设您有这段代码(在 c++14 中):

auto && a = a_function_returning_a_temp_rvalue();

返回值存储在哪里，何时调用其析构函数，是a const(考虑到函数不返回const ), 标准中是如何规定的？ (这是否合法)

此外，您是否确认临时对象确实绑定(bind)到 a，并且不会调用 operator= 这两个构造函数？

它在 g++ 上编译并避免键入长模板类，但在使用它之前，我想检查它是否在 c++14 中明确说明

Answer 1

规则在[class.temporary]中:

There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression. [...] The third context is when a reference is bound to a temporary.116 The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:
— A temporary object bound to a reference parameter in a function call (5.2.2) [...]
— The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is [...]
— A temporary bound to a reference in a new-initializer (5.3.4) persists until [...]

在这个例子中:

auto && a = a_function_returning_a_temp_rvalue();

我假设该函数返回类型为 T 的内容(与 T& 或 T&& 相对)。在这种情况下，我们确实有一个临时绑定(bind)到一个引用，并且这些异常(exception)都不适用。因此，该临时对象的生命周期延长了 a 的生命周期。