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Can a class member function be invoked without an object?

Time:01-18

I was learning the history about Lambda's in C and saw the following code (which is not lambda) but I am surprised how it Works

struct Printer{
void operator() (int x) const{
    std::cout << x << '\n';
 }
};

 int main(){
   std::vector <int> vint;
   //doing it the C   03 way
    vint.push_back(1);
    vint.push_back(7);

 std::for_each(vint.begin(),vint.end(), Printer());

}

How is the Printer() call in the for_each function working?

CodePudding user response:

Printer() is an instance of the Printer class. It will result in a temporary object of type Printer which is passed to std::for_each.

This is the object on which operator() is called by std::for_each internally.

Without an object of type Printer, it is not possible to call the operator() member function.

CodePudding user response:

Can a class member function be invoked without an object?

If it is a static member function then yes, it can be invoked without instance of the class.

If it is a non-static member function, then no, it can not be invoked without instance of the class. The example operator() function is a non-static member function, so it cannot be called without an instance.

How is the Printer() call in the for_each function working?

Printer() is syntax for value initialisation of a temporary object. The temporary instance of Printer class is passed as an argument into std::for_each.

CodePudding user response:

If the std::for_each is confusing (you provide a callable which is called by the algorithm), replace it with

 for (auto i: vint) Printer()(i);

which does the same thing: make a temporary object and call operator() with the needed parameter.

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