在这种情况下会发生什么？
// assume WeakPtr is valid and has not expired
const auto& something = WeakPtr.lock();
something->doStuff();
这是未定义的吗？
在这种情况下它会改变吗？
std::shared_ptr<Something> getSomething() { return mSomething.lock(); }
const auto& something = getSomething();
那这个呢？
std::vector<int> getInts() { return std::vector<int>{ 1, 2, 3 }; }
const auto& ints = getInts();
在每一种情况下，const auto& 都暗示我想绑定(bind)一个对象的引用，但在每一种情况下，我都将它绑定(bind)到一个临时的右值对象。我是在招来灾难吗？
最佳答案
Is this undefined?
每个案例都有明确的定义。
What happens in this case?
在每种情况下，临时对象的生命周期都会延长，以匹配标准 [class.temporary] 部分中描述的 const 引用的生命周期。
[class.temporary] (standard draft)
4 There are two contexts in which temporaries are destroyed at a different point than the end of the full- expression. The first context is ... [irrelevant to your cases]
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 ... [a few exceptions which do not apply to your cases]
关于c++ - 将右值分配给 'const auto&' 时会发生什么，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42868744/