c++ - 返回临时的 const 引用

Question

为什么函数foo给出警告“returning reference to temporary”

const bool& foo()
{
    return true;
}

如果像这样声明 bar 是好的并且不会产生任何类型的警告

const bool& bar = true;

PS: 我正在使用 GCC

Answer 1

第二个很好，因为 lifetime of temporary bool 由 true 构造，将延长引用 bar 的生命周期。

Whenever a reference is bound to a temporary object or to a subobject thereof, the lifetime of the temporary object is extended to match the lifetime of the reference

但是生命周期规则不适用于第一种情况。

There are following exceptions to this lifetime rule:

• a temporary bound to a return value of a function in a return statement is not extended: it is destroyed immediately at the end ofthe return expression. Such return statement always returns a danglingreference.

所以 foo 总是返回一个悬空引用。