What is the meaning of bool operator () (int num)?-CodePudding

Is () used in the bool operator()(int num) of the below code a another operator or it means something else ? Please help.

#include <iostream>
#include <vector>
#include<algorithm>
using namespace std;

class IsOdd{
    public:
        bool operator()(int num){
            return ((num%2)==1);
        }
};
int main(){

    vector <int> v {1,2,3,4,5,6,7,8,9,10};
    vector <int>::iterator pend;
    vector<int>:: iterator q;

    pend = remove_if(v.begin(), v.end(), IsOdd());
    for(q=v.begin(); q!=pend;  q)
        cout<<*q<<endl;
    return 0;
}

CodePudding user response：

That means overloading the operator () to work with the IsOdd class.

Here is a sample of code of the overloaded operator in action:

#include <iostream>
#include <vector>
#include<algorithm>

class IsOdd {
public:
    bool operator()(int num) {
        return ((num % 2) == 1);
    }
};

int main()
{
    std::vector <int> v{ 1,2,3,4,5,6,7,8,9,10 };

    IsOdd isOdd; // Defining object isOdd of the class IsOdd

    for (int i = 0; i < v.size(); i  )
    {
        if (isOdd(v[i])) std::cout << v[i] << " is odd." << std::endl;
        else std::cout << v[i] << " is not odd." << std::endl;
    }
    return 0;
}

I've also removed the iterators and shown with a much simpler way to iterate through elements in a std::vector.

Also, you should not use the following in your code:

using namespace std;

...as it's considered as bad practice.

Output:

1 is odd.
2 is not odd.
3 is odd.
4 is not odd.
5 is odd.
6 is not odd.
7 is odd.
8 is not odd.
9 is odd.
10 is not odd.

CodePudding user response：

It's a function, and the () is part of the name - operator() is a special spelling for "function that is used to make an object behave like a function".
(It's a bit strange to call it "operator", but the C committee prefers reusing keywords to introducing new ones.)

If you have an object of the IsOdd type, such as

IsOdd is_odd;

then what looks like a regular function call,

is_odd(3)

is transformed behind the scenes into

is_odd.operator() (3)

CodePudding user response：

IsOdd is a functor class. It has an overload of the function call operator, so that given an instance x you can call it via x(42) to determine if 42 is odd.

In the code it is used to pass a callable to the algorithm. Nowadays you would rather use a lambda:

    auto isOdd = [](int x) { return ((num % 2) == 1);};
    pend = remove_if(v.begin(), v.end(), isOdd);

or perhaps, if the functor is only needed for the algorithm but not elsewhere, define the lambda in-line:

    pend = remove_if(v.begin(), v.end(), [](int x){ return ((num % 2) == 1);});

Is () used in the bool operator()(int num) of the below code a another operator or it means something else ?

Here:

pend = remove_if(v.begin(), v.end(), IsOdd());

the () in IsOdd() is a call to the constructor to create an instance of the class IsOdd. The operator() of that instance is then called by the algorithm. If you like you could also call it yourself:

auto x = IsOdd();    // call constructor
std::cout << x(42);  // call operator()

CodePudding user response：

It means it's a function named operator(), returns a boolean, and takes in an integer named num.

Edit: Its an overloaded function operator so you can have it take no arguments or the argument int num.