如何检查一个元素是否在集合中?

是否有与以下代码更简单的等效代码:

myset.find(x) != myset.end()

当前回答

从c++ 20开始,就有了bool std::contains(const K&) https://en.cppreference.com/w/cpp/container/set/contains

其他回答

/ /通用语法

       set<int>::iterator ii = find(set1.begin(),set1.end(),"element to be searched");

/*在下面的代码中,我试图找到元素4和int集,如果它存在与否*/

set<int>::iterator ii = find(set1.begin(),set1.end(),4);
 if(ii!=set1.end())
 {
    cout<<"element found";
    set1.erase(ii);// in case you want to erase that element from set.
 }

就是这个,一英里。

bool once(uintptr_t val) {
    return visited.emplace(val).second;
}

怎么会不是这样呢?

https://godbolt.org/z/9zP77jqMc

func5(unsigned long):
        sub     rsp, 24
        mov     QWORD PTR [rsp+8], rdi
        lea     rsi, [rsp+8]
        mov     edi, OFFSET FLAT:visited2
        call    std::pair<std::_Rb_tree_iterator<unsigned long>, bool> std::_Rb_tree<unsigned long, unsigned long, std::_Identity<unsigned long>, std::less<unsigned long>, std::allocator<unsigned long> >::_M_emplace_unique<unsigned long&>(unsigned long&)
        add     rsp, 24
        mov     eax, edx
        ret

Just to clarify, the reason why there is no member like contains() in these container types is because it would open you up to writing inefficient code. Such a method would probably just do a this->find(key) != this->end() internally, but consider what you do when the key is indeed present; in most cases you'll then want to get the element and do something with it. This means you'd have to do a second find(), which is inefficient. It's better to use find directly, so you can cache your result, like so:

auto it = myContainer.find(key);
if (it != myContainer.end())
{
    // Do something with it, no more lookup needed.
}
else
{
    // Key was not present.
}

当然,如果你不关心效率,你总是可以自己滚动,但在这种情况下,你可能不应该使用c++…;)

检查许多STL容器是否存在的典型方法,如std::map, std::set,…是:

const bool is_in = container.find(element) != container.end();

您还可以在插入元素时检查元素是否在set中。 单元素版本返回一个pair,其成员pair::first set指向一个迭代器,该迭代器要么指向新插入的元素,要么指向集合中已经存在的等效元素。如果插入了新元素,pair中的第二个元素将被设置为true,如果已经存在等效元素则为false。

例如:假设集合中已经有20作为元素。

 std::set<int> myset;
 std::set<int>::iterator it;
 std::pair<std::set<int>::iterator,bool> ret;

 ret=myset.insert(20);
 if(ret.second==false)
 {
     //do nothing

 }
 else
 {
    //do something
 }

 it=ret.first //points to element 20 already in set.

如果元素是新插入的,则than pair::first将指向新元素在set中的位置。