How to check that an element is in a std::set?

Question

How do you check that an element is in a set?

Is there a simpler equivalent of the following code:

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

Answer 1

//general Syntax

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

/* in below code i am trying to find element 4 in and int set if it is present or not*/

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.
 }

Answer 2

In C++20 we'll finally get std::set::contains method.

#include <iostream>
#include <string>
#include <set>

int main()
{
    std::set<std::string> example = {"Do", "not", "panic", "!!!"};

    if(example.contains("panic")) {
        std::cout << "Found\n";
    } else {
        std::cout << "Not found\n";
    }
}

Answer 3

The typical way to check for existence in many STL containers such as std::map, std::set, ... is:

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

Answer 4

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.
}

Of course, if you don't care about efficiency, you can always roll your own, but in that case you probably shouldn't be using C++... ;)