Using "std::find" from the algorithm librairy for a list of struct?

hoboCPP (10)
Hi, I'm trying to write a method that returns true or false whether this element exists in my list.

Basically, I have a list of a struct. 

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struct Person
{
	int age;
	string name;
};


I have a list of persons.
 
 
std::list<Person> listPerson;

and I have the following method:
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bool personExists(int age) const
	{

		auto list= listPerson;
		auto it = find(list.begin(), list.end(), ?????);
		return it != list.end();
	}

At the ?????, I just want to check if in the list, there is a Person with x age.
Is there a way to use the std::find to do that or should I just manually code a for loop with a if statement?

Thank you.
Last edited on
keskiverto (10399)
No and no.

The std::find uses a value. That is not convenient.

No need for loop, use std::find_if
http://www.cplusplus.com/reference/algorithm/find_if/
gunnerfunner (2127)
std::find() uses operator == for comparison, so if your question specifically asked you to use std::find() and std::find() only - you could, by overloading this operator. Otherwise std::find_if() is certainly the way to go: 
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# include <iostream>
# include <algorithm>
# include <list>
# include <string>

struct Person
{
    int m_age;
    std::string m_name;

    bool operator == (const int age)const
    {
        return m_age == age;
    }
    Person (const int age, const std::string& name)
        : m_age(age), m_name(name){}
};


int main()
{
    std::list<Person> personsList{};
    personsList.emplace_back(42, "john");
    personsList.emplace_back(24, "jane");

    if (std::find(personsList.cbegin(), personsList.cend(), 42) != personsList.cend())
    {
        std::cout << "found \n";
    }
    else
    {
        std::cout << "not found \n";
    }
}

hoboCPP (10)
Oh thanks guys. I didn't look into the find_if.
Cheers!"
dhayden (5798)
auto list= listPerson;
I believe this makes a copy of the list. That would be very inefficient.
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