pop_back（）返回值？ [英] pop_back() return value?

问题描述

为什么 pop_back（）有返回值？我有谷歌这方面，并发现，它使它更有效率。

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

我认为有一些事情涉及到复制最后一个对象的实例可能会抛出异常。当这样做，你失去了你的对象，因为pop_back（）删除它从你的容器。最好用几行代码：

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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