STL删除不能按预期工作？ [英] STL remove doesn't work as expected?

问题描述

int main()
{

        const int SIZE = 10;
        int a[SIZE] = {10, 2, 35, 5, 10, 26, 67, 2, 5, 10};
        std::ostream_iterator< int > output(cout, " ");
        std::vector< int > v(a, a + SIZE);
        std::vector< int >::iterator newLastElement;

        cout << "contents of the vector: ";
        std::copy(v.begin(), v.end(), output);

        newLastElement = std::remove(v.begin(), v.end(), 10);
        cout << "\ncontents of the vector after remove: ";
        //std::copy(v.begin(), newLastElement, output); 
                         //this gives the correct result : 2 35 5 26 67 2 5
        std::copy(v.begin(), v.end(), output);
          //this gives a 10 which was supposed to be removed : 2 35 5 26 67 2 5 2 5 10

        cout << endl;
        return 0;
}

数组中有三个10。

为什么在我们用remove函数删除所有10个数组后，数组v包含10。

why does the array v contains a 10 after we remove the all the 10s with remove function.

a href =http://ideone.com/rlxd8>此处

you can see the compiled output also here

推荐答案

实际上 std :: remove 不会从容器中删除该项目。引用自此处

Actually std::remove doesn't remove the item from the container. Quoted from here

删除 [第一个，最后）范围内所有等于 value 。也就是说，remove返回一个迭代器 new_last ，使得范围 [first，new_last] 不包含等于 value 。 [new_last，last）的迭代器都是仍可解除引用，但它们指向的元素未指定。删除是稳定的，意味着不等于值的元素的相对顺序不变。

Remove removes from the range [first, last) all elements that are equal to value. That is, remove returns an iterator new_last such that the range [first, new_last) contains no elements equal to value. The iterators in the range [new_last, last) are all still dereferenceable, but the elements that they point to are unspecified. Remove is stable, meaning that the relative order of elements that are not equal to value is unchanged.`

也就是说， std :: remove 仅适用于一对迭代器，并且不知道实际包含项目的容器的任何内容。事实上， std :: remove 不可能知道底层容器，因为没有办法从一对迭代器去发现容器迭代器属于。因此 std :: remove 并不真正删除项目，只是因为它不能。从 移除容器中的项目的唯一方法是调用该容器上的成员函数。

That is, std::remove works with a pair of iterators only and does not know anything about the container which actually contains the items. In fact, it's not possible for std::remove to know the underlying container, because there is no way it can go from a pair of iterators to discover about the container to which the iterators belong. So std::remove doesn't really remove the items, simply because it cannot. The only way to actually remove an item from a container is to invoke a member function on that container.

项目，然后使用擦除删除成语：

 v.erase(std::remove(v.begin(), v.end(), 10), v.end()); 

---

erase-remove idiom 是如此常见和有用的是 std :: list 添加了另一个名为 list :: remove 的成员函数，它产生的效果与 erase-remove idiom。

---

The erase-remove idiom is so common and useful is that std::list has added another member function called list::remove which produces the same effect as that of the erase-remove idiom.

 std::list<int> l;
 //...
 l.remove(10); //it "actually" removes all elements with value 10!

这意味着你不需要使用 erase-remove idiom当你使用 std :: list 。您可以直接调用其成员函数 list :: remove 。

That means, you don't need to use erase-remove idiom when you work with std::list. You can directly call its member function list::remove.

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