STL“擦除 - 去除” idiom：为什么不“调整大小 - 删除”？ [英] STL "erase-remove" idiom: Why not "resize-remove"?

问题描述

一般来说，从 std :: vector 中完全删除所需项目的好方法是 erase-remove idiom 。

It is commonly understood that a good way to fully delete desired items from a std::vector is the erase-remove idiom.

如上述链接中所述（截至本发布日期）代码 erase-remove 成语如下：

As noted in the above link (as of the date of this posting), in code the erase-remove idiom looks like this:

int main()
{
  // initialises a vector that holds the numbers from 0-9.
  std::vector<int> v = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

  // erase-remove idiom to completely eliminate the desired items from the vector
  v.erase( std::remove( std::begin(v), std::end(v), 5 ), std::end(v) ); 
}

我想知道 resize-remove idiom在功能和性能方面等同于 erase-remove idiom。

I would like to know whether a resize-remove idiom is equivalent in terms of functionality and performance to the erase-remove idiom. Or, perhaps I am missing something obvious?

以下是 resize-remove idiom等同于上述 erase-remove idiom？

Is the following resize-remove idiom equivalent to the above erase-remove idiom?

int main()
{
  std::vector<int> v = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

  // Is this "resize-remove" approach equivalent to the "erase-remove" idiom?
  v.resize( std::remove( std::begin(v), std::end(v), 5 ) - v.begin() ); 
}

推荐答案

两个原因：

1. std :: remove ， - op需要随机访问迭代器。

1. std::remove algorithm requires only Forward Iterator, but - op requires Random Access Iterator.

std :: remove 表示容器的新端。逻辑上，我们应该清除[容器的新端，容器的旧端）。

The result of std::remove means "the new end of container". Logically, we should erase [ "the new end of container" , "the old end of container" ).

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