pop_back() 返回值?

pop_back() return value?(pop_back() 返回值?)

问题描述

为什么 pop_back() 没有返回值?我在谷歌上搜索过这个，发现它更有效.这是在标准中这样做的唯一原因吗?

Why doesn't pop_back() have a return value? I have Googled regarding this and found out that it makes it more efficient. Is this the only reason for making it so in the standard?

推荐答案

我认为复制最后一个对象的实例可能会引发异常这一事实与此有关.这样做时，您将丢失对象，因为 pop_back() 确实将其从容器中删除.用几行代码更好:

I think there is something related to the fact that copying an instance of the last object could throw an exception. When doing so, you're losing your object, since pop_back() did remove it from your container. Better with a few lines of code:

std::vector<AnyClass> holds = {...} ;
try {
  const AnyClass result = holds.pop_back(); // The copy Ctor throw here!
} catch (...)
{ 
 // Last value lost here. 
}

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