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++…；）

另一种简单地判断元素是否存在的方法是检查count()

if (myset.count(x)) {
   // x is in the set, count is 1
} else {
   // count zero, i.e. x not in the set
}

然而，大多数时候，我发现自己需要访问元素，无论我在哪里检查它的存在。

所以我还是要找到迭代器。当然，最好也把它和end进行比较。

set< X >::iterator it = myset.find(x);
if (it != myset.end()) {
   // do something with *it
}

C + + 20

在c++ 20中，set获得一个contains函数，因此如下所述成为可能:https://stackoverflow.com/a/54197839/895245

if (myset.contains(x)) {
  // x is in the set
} else {
  // no x 
}

我能够为std::list和std::vector编写一个通用的包含函数，

template<typename T>
bool contains( const list<T>& container, const T& elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

template<typename T>
bool contains( const vector<T>& container, const T& elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

// use:
if( contains( yourList, itemInList ) ) // then do something

这样可以稍微清理一下语法。

但是我不能使用模板模板参数魔术使此工作任意stl容器。

// NOT WORKING:
template<template<class> class STLContainer, class T>
bool contains( STLContainer<T> container, T elt )
{
  return find( container.begin(), container.end(), elt ) != container.end() ;
}

任何关于改进上一个答案的评论都是很好的。

写你自己的:

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

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

/ /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.
 }