删除循环内向量的元素 [英] Remove elements of a vector inside the loop

问题描述

我知道有与这个类似的问题，但我没能在他们的帮助下找到我的代码的方法.我只想通过检查循环内此元素的属性来删除/删除向量的元素.我怎样才能做到这一点?我尝试了以下代码，但收到了模糊的错误消息:

I know that there are similar questions to this one, but I didn’t manage to find the way on my code by their aid. I want merely to delete/remove an element of a vector by checking an attribute of this element inside a loop. How can I do that? I tried the following code but I receive the vague message of error:

'operator =' 功能在播放器"中不可用.

'operator =' function is unavailable in 'Player’.

 for (vector<Player>::iterator it = allPlayers.begin(); it != allPlayers.end(); it++)
 {
     if(it->getpMoney()<=0) 
         it = allPlayers.erase(it);
     else 
         ++it;
 }

我该怎么办?

更新:你认为问题vector::erase with pointer成员属于同样的问题?因此我需要赋值运算符吗?为什么?

Update: Do you think that the question vector::erase with pointer member pertains to the same problem? Do I need hence an assignment operator? Why?

推荐答案

你不应该在 for 循环中增加 it :

You should not increment it in the for loop:

for (vector<Player>::iterator it=allPlayers.begin(); 
                              it!=allPlayers.end(); 
                              /*it++*/) <----------- I commented it.
{

   if(it->getpMoney()<=0) 
      it = allPlayers.erase(it);
  else 
      ++it;
 }

注意注释部分；it++ 在那里不需要，因为 it 在 for-body 本身中递增.

Notice the commented part;it++ is not needed there, as it is getting incremented in the for-body itself.

至于'operator =' function isavailable in 'Player'"的错误，来自于内部使用的erase()的使用operator= 移动向量中的元素.为了使用erase()，Player类的对象必须是可赋值的，这意味着你需要为实现operator=播放器类.

As for the error "'operator =' function is unavailable in 'Player’", it comes from the usage of erase() which internally uses operator= to move elements in the vector. In order to use erase(), the objects of class Player must be assignable, which means you need to implement operator= for Player class.

无论如何，您应该避免原始循环¹ 尽可能多，应该更喜欢使用算法.在这种情况下，流行的 Erase-Remove Idiom 可以简化您的操作正在做.

Anyway, you should avoid raw loop¹ as much as possible and should prefer to use algorithms instead. In this case, the popular Erase-Remove Idiom can simplify what you're doing.

allPlayers.erase(
    std::remove_if(
        allPlayers.begin(), 
        allPlayers.end(),
        [](Player const & p) { return p.getpMoney() <= 0; }
    ), 
    allPlayers.end()
); 

^{1.这是我看过的Sean Parent 的最佳演讲之一.}

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