C ++标准库是否提供更紧凑，更通用的“擦除删除"习惯用法版本? [英] Does C++ standard library provide more compact and generalized version of the erase–remove idiom?

问题描述

我们可以通过流行的擦除–从容器中删除一个元素/条目–删除成语. 但是，在应用此惯用语时，我们许多人会遇到一些问题:

We can erase one element/ entry from a container by the popular erase–remove idiom. However, many of us would have encountered some problems while applying this idiom:

• 一个人很容易陷入 typos 之类的陷阱

c.erase(std::remove_if(c.begin(), c.end(), pred));
//                                             , c.end() //---> missing here

或

c.erase((std::remove_if(c.begin(), c.end(), pred), c.end()))
//      ^^                                               ^^
// extra () makes it pass only c.end() to the c.erase

对于诸如此类的容器，它甚至遵循错误的语义 std::list通过不选择其自己的成员 std::list::remove_if() 为成语.

第三，使用std::remove_if 不适用于关联 容器 .

It even follows the wrong semantics for containers like std::list by not selecting its own member std::list::remove_if() for the idiom.

Thirdly, using std::remove_if does not work for associative containers.

在std::erase-std::remove_if或std::erase_if之类的更通用且更不易错字的内容? class ="post-tag" title =显示标记为'c ++ 17'"的问题rel ="tag"> c ++ 17 ，否则

Do we have anything generalized and less typo-prone than std::erase-std::remove_if or something like std::erase_if within the scope of c++17, or will there be such a utility in c++20?

推荐答案

不在 n4009论文中提到了容器一致性擦除的建议，并最终在 C ++ 20标准为 std::erase_if 每个容器的非成员功能.

Not in the scope of c++17, but c++20 onwards!

Yes. The proposal of consistent container erasure has been mentioned in n4009 paper and finally adopted in C++20 standard as std::erase_if which is a non-member function for each containers.

这确保了std::basic_string和所有标准容器的统一容器擦除语义，除了std::array(因为它具有固定大小).

This ensures a uniform container erasure semantics for std::basic_string and all standard containers, except std::array(as it has the fixed-size).

这意味着样板代码

container.erase(
    std::remove_if(
        container.begin(), container.end(),
        [](const auto& element) ->bool { return /* condition */; }),
    vec.end());

将简单地分解为

std::erase_if(container, [](const auto& element) ->bool { return /* condition */; });

第二，此统一语法为每个容器选择正确的语义.这意味着

Secondly, this uniform syntax selects the proper semantics for each container. This means

• 对于诸如 std::vector ， std::deque 并用于 std::std::basic_string ，它等效于

• For sequence containers like std::vector, std::deque and for std::std::basic_string, it will be equivalent to

container.erase(
       std::remove_if(container.begin(), container.end(), unaryPredicate)
       , container.end()
);

对于序列容器 std::forward_list 和 std::list ， 等同于

For sequence containers std::forward_list and std::list, it will be equivalent to

container.remove_if(unaryPredicate);

对于有序关联容器(即 std::set ， std::map ， std::multiset 和 std::unordered_set ， std::unordered_multiset 和

For ordered associative containers(i.e. std::set, std::map, std::multiset and std::multimap) and unordered associative containers(i.e. std::unordered_set, std::unordered_map, std::unordered_multiset and std::unordered_multimap), the std::erase_if is equivalent to

for (auto i = container.begin(), last = container.end(); i != last; ) 
{
  if (unaryPredicate(*i)) 
  {
    i = container.erase(i);
  }
  else
  {
    ++i;
  }
}

除此之外，该标准还为表单的序列容器添加了 std::erase

In addition to that, the standard also added std::erase for sequence containers of the form

std::erase(container, value_to_be_removed);

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