C++ function template initialization?

Question

I'm not sure if this title reflects the question. Here's a template function. My question is

what does s(...) in the code below mean since compiler doesn't know what class Something is and the compiler doesn't even complain.

I don't quite understand what's going on. Thanks

template <class Something>
void foo(Something s)
{
    s(0, {"--limit"}, "Limits the number of elements.");  //What does this mean?
    
    s(false, {"--all"}, "Run all");
}

Try to call `foo` with some type (e.g. `foo(5)` which will determine `Something` to be an `int`) and the compiler will instantiate it and issue an error if needed. — wohlstad, Aug 26 '22 at 05:51
Stackoverflow is not an *"introduction to C++"*. This is explained in any beginner level [c++ book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). Read about *1)* what are function templates *2)* what are functors *3)* what is template argument deduction *4)* what is operator overloading — Jason, Aug 26 '22 at 05:59

score 1 · Answer 1 · answered Aug 26 '22 at 06:05

Looks like the intention of foo is to accept a class that has a call operator.

Eg

#include <iostream>
#include <string>
#include <vector>

struct ReallySomething {
    void operator()(bool b, const std::vector<std::string>& flags, const std::string& description)
    {
        std::cout << "Condition is " << (b ? "true" : "false") << std::endl;
        std::cout << "Number of flags: " << flags.size() << std::endl;
        std::cout << "Description: " << description << std::endl;
    }
};

template <class Something>
void foo(Something s)
{
    s(0, {"--limit"}, "Limits the number of elements.");  //What does this mean?
    
    s(false, {"--all"}, "Run all");
}

int main()
{
    ReallySomething r;
    foo(r);

    return 0;
}

Then

s(false, {"--all"}, "Run all");

and

s(0, {"--limit"}, "Limits the number of elements.");

will call this function.

Try the code here

score 1 · Accepted Answer · answered Aug 26 '22 at 06:08

what does s(...) in the code below mean

It means that you're trying to call s while passing different arguments 0, {"--limit"}, "Limits the number of elements. For example, s might be of some class-type that has overloaded the call operator operator() that takes variable number of arguments.

So by writing:

s(0, {"--limit"}, "Limits the number of elements."); //you're using the overloaded call operator() for some given class type

In the above, you're using the overloaded call operator() and passing different arguments to it.

Lets look at a contrived example:

struct Custom 
{
  //overload operator()
  template<typename... Args> 
  void operator()(const Args&...args)
  {
      std::cout << "operator() called with: "<<sizeof...(args)<<" arguments"<<std::endl;
  }
};

template <class Something>
void foo(Something s)
{
    s(0, "--limit", "Limits the number of elements.");  //call the overloaded operator()
    
    s(false, "--all", "Run all", 5.5, 5);              //call the overloaded operator()
}
int main()
{
    Custom c;      //create object of class type Custom
    foo(c);       //call foo by passing argument of type `Custom`
}

Demo

C++ function template initialization?

2 Answers2