Thread: Template function as for_each argument

Code:

/*
 * Practice.cpp
 *
 *  Created on: Sep 25, 2011
 *      Author: mike
 */
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

template <class T>
void printElement(const T& element) {

	cout << element << ' ';
}

class OutputFunctor {
public:
  template <class T>
  void operator()(const T& val)  { printElement(val); }
};


int main() {


	vector<int> nums1;

	for (int i = 1; i <= 10; ++i)
		nums1.push_back(i);


        // Compiler throws a fit
//	for_each (nums1.begin(), nums1.end(), printElement);
//		cout << '\n';

	// Not a problem  
	for_each (nums1.begin(), nums1.end(), OutputFunctor());
	cout << '\n';


	return 0;
}

How come the compiler chokes on a template function, but greedily gobbles up my function object containing the same function?

Originally Posted by wildcard_seven

How come the compiler chokes on a template function, but greedily gobbles up my function object containing the same function?

You did not instantiate your function template, e.g.,

Code:

for_each (nums1.begin(), nums1.end(), printElement<int>);

Originally Posted by wildcard_seven

It's not like you have to mention the type during a regular call to a template function.

But you do, if the template argument cannot be deduced from the arguments, or if you want to provide a different template argument

Originally Posted by laserlight

But you do, if the template argument cannot be deduced from the arguments, or if you want to provide a different template argument

In the second for_each in the original post, the compiler can just guess the type of OutputFunctor to create based on the iterators provided ?
If so, why can't it do so for the first one ?

That argument can be of any type. In the code of for_each the argument is called as a function/function object. That's where the argument for operator() of OutputFunctor is deduced.

The problem is that printElement is not a type, it's the name of a function template. Functions have addresses, function templates don't. Function templates need to be instantiated before they become functions.

(Your terminology is a bit wrong. printElement is not a template function = a special kind of function. It is a function template = a particular kind of template.)

Another way to look at it:
for_each takes a type T, but your function is of type template<typename K> (IIRC) (that is, a type describing a type that takes a type). Type mismatch, and no implicit conversion is possible. You have to explicitly specify that type K in order to create a proper type. I forget the actual terms for these things, but there you have it.

For the function object, you are creating a new object which obviously is a type, which allows the compiler to automatically deduce T. Inside for_each, the () operator is called and parameters are passed in. Hence, the template parameter for the operator is automatically deduced as well.