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++…；）

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++…；）

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

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。

我使用

if(!my_set.count(that_element)) //Element is present...
;

但它的效率不如

if(my_set.find(that_element)!=my_set.end()) ....;

我的版本只是节省了我写代码的时间。对于竞争性编码，我更喜欢这种方式。

写你自己的:

template<class T>
bool checkElementIsInSet(const T& elem, const std::set<T>& container)
{
  return container.find(elem) != container.end();
}