引用函数的 return 值

Question

我很困惑在下面的代码片段中是否保证 foo 有效，即是否允许我将函数 return 值存储为 const 引用？

我问是因为 return 值存储在堆栈框架中，该值在 returned 后不久可能无效。

Foo getFoo() {
   return Foo();
}

void bar() {
  const auto& foo = getFoo();

  // more function calls...

  // is foo guaranteed to be valid?
}

Answer 1

它是有效的 C++，但请注意 一些编译器（据我所知，例如 Visual Studio 2015）不实现它正确.

根据标准(N4140):

12.2 Temporary objects
....

⁴ 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. If the constructor has one or more default arguments, the destruction of every temporary created in a default argument is sequenced before the construction of the next array element, if any.

⁵ 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 subobject to which the reference is bound persists for the lifetime of the reference except:

(some exceptions)

引用函数的 return 值

Reference to return value of function

c++

reference

return-value

stack-frame