擦除和删除之间的区别 [英] Difference between erase and remove

问题描述

我对使用std :: remove算法之间的区别感到困惑。具体来说，我不能理解什么是当我使用这种算法被删除。我写了一个小测试代码，像这样：

  std :: vector< int&一个; 
 a.push_back（1）; 
 a.push_back（2）; 
 
 std :: remove（a.begin（），a.end（），1）; 
 
 
 int s = a.size（）; 
 
 std :: vector< int> :: iterator iter = a.begin（）; 
 std :: vector< int> :: iterator endIter = a.end（）; 
 
 std :: cout<<Using iter ... \\\
; 
 for（; iter！= endIter; ++ iter）
 {
 std :: cout<< * iter<<\\\
; 
} 
 
 std :: cout<<Using size ... \\\
; 
 for（int i = 0; i  {
 std :: cout<< a [i]< \\ n; 
} 
  

在这两种情况下，输出为2,2。 b
$ b

但是，如果我使用erase删除这样的东西：

  a.erase （std :: remove（a.begin（），a.end（），1），a.end（））; 
  

我得到的结果为2。

所以我的问题是：

（1）。是否使用std :: remove除了使用它与擦除功能。

（2）。即使在做了std :: remove之后，为什么a.size（）返回2而不是1？

我在Scott Meyer的有效STL书中阅读了有关erase-remove成语。

解决方案

remove（） t实际上从容器中删除元素 - 它只是在删除的元素的顶部分流未删除的元素。关键是要认识到 remove（）被设计为不只是一个容器，而是任何任意的前向迭代器对：这意味着它不能实际删除元素，因为任意的迭代器对不一定具有删除元素的能力。

例如，指针正则C数组的开始和结束是前向迭代器，因此可以使用 remove（）：

  int foo [100]; 
 
 ... 
 
 remove（foo，foo + 100，42）; //删除所有等于42的元素
  

无法调整数组大小！

I am bit confused about the difference between the usage of std::remove algorithm. Specifically I am not able to understand what is being removed when I use this algorithm. I wrote a small test code like this:

std::vector<int> a;
a.push_back(1);
a.push_back(2);

std::remove(a.begin(), a.end(), 1);


int s = a.size();

std::vector<int>::iterator iter = a.begin();
std::vector<int>::iterator endIter = a.end();

std::cout<<"Using iter...\n";
for(; iter != endIter; ++iter)
{
	std::cout<<*iter<<"\n";
}

std::cout<<"Using size...\n";
for(int i = 0; i < a.size(); ++i)
{
	std::cout<<a[i]<<"\n";
}

The output was 2,2 in both the cases.

However, if I use erase with the remove something like this:

a.erase(std::remove(a.begin(), a.end(), 1), a.end());

I get the output as 2.

So my questions are:

(1). Is there any use of std::remove other than using it with erase function.

(2). Even after doing std::remove, why a.size() returns 2 and not 1?

I read the item in Scott Meyer's Effective STL book about the erase-remove idiom. But am still having this confusion.

解决方案

remove() doesn't actually delete elements from the container -- it only shunts non-deleted elements forwards on top of deleted elements. The key is to realise that remove() is designed to work on not just a container but on any arbitrary forward iterator pair: that means it can't actually delete the elements, because an arbitrary iterator pair doesn't necessarily have the ability to delete elements.

For example, pointers to the beginning and end of a regular C array are forward iterators and as such can be used with remove():

int foo[100];

...

remove(foo, foo + 100, 42);    // Remove all elements equal to 42

Here it's obvious that remove() cannot resize the array!

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