...is not a specialization of a function template

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When I try to compile the following code I get an error about the specialization.

C++

#include <algorithm>
#include <time.h>
#include <stdlib.h>
#include <stdio.h>
#include <iostream>

using namespace std;

class MyRandom {
public:
	MyRandom() {
		srand(time(0));
	}

	int operator() () const {
		return rand() % 10;
	}
};

template<typename T>
class MyPrint {
public:
	void operator() (T const& v) const {
		cout << v << endl;
	}

	template<>
	void operator() (int const& v) const {
		cout << "int:" << v << endl;
	}
};


int main() {
	int elements[10];

	generate(elements, elements + 10, MyRandom());

	for_each(elements, elements + 10, MyPrint<int>());

	return 0;
}

I don't get it, it seems ok to me :(

Posted 16-May-14 0:13am

Sergejack

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2 solutions

Solution 2

The member function of a templated class cannot shed the class template parameter, which appears to be your intention. The compiler in fact treats the member function as a non-templated function within the specific instantiation of the templated class for a given class template argument. I. e., if you never instantiate the calss template, the compiler won't even notice, but if you do, the argument passed to the class template instantiation will be used and can't be a free template argument for member functions.

E. g. if you have the following code:

C++

template <class t>
class myT {
   T var;
public:
   void hello() const;
};
void foo {
   myT<int> x;
   x.hello();
}

Then the function hello will be instantated as myT<int>::hello().

Posted 16-May-14 0:32am

Stefan_Lang

Updated 16-May-14 0:34am

v3

Comments

Sergejack 16-May-14 7:54am

I don't really follow you. Could you explain it with other words?

Stefan_Lang 16-May-14 8:15am

You can declare template classes and template functions. Your class MyPrint is declared as a template class. When you used a template syntax for your member function operator() however, that was template function syntax, and the argument list defined therein only applies to the function.

If you define anything within the template class definition, you always refer to the (unspecified) template arguments. If you want to refer to specific values of a template argument, you need to make a separate definition, like CPallini did in his solution.

I wasn't even aware that was possible, but CPallini did the right thing: instantiate the member function for the specialized class.

CPallini 16-May-14 8:02am

My 5. By the way, many thanks for having made the damned CodeProject 'styler' work on my answer.

Stefan_Lang 16-May-14 8:24am

Thanks, and yw!

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CPallini · Accepted Answer · 2014-05-16T00:30:00

Solution 1

Try:

C++

#include <algorithm>
#include <time.h>
#include <stdlib.h>
#include <stdio.h>
#include <iostream>

using namespace std;

class MyRandom {
public:
  MyRandom() {
    srand(time(0));
  }

  int operator() () const {
    return rand() % 10;
  }
};

template<typename T>
class MyPrint {
public:
  void operator() (T const& v) const {
    cout << v << endl;
  }
};

template <> void MyPrint<int>::operator()(int const &v) const { cout << "int: " << v << endl;}


int main() {
  int elements[10];

  generate(elements, elements + 10, MyRandom());

  for_each(elements, elements + 10, MyPrint<int>());

  return 0;
}

C++

Posted 16-May-14 0:30am

CPallini

Updated 16-May-14 1:28am

v2

Comments

Sergejack 16-May-14 7:51am

It works. So I can specialize a function template from inside the class declaration?

CPallini 16-May-14 7:59am

I ma not an expert, but apparently you cannot (and it makes sense).

Sergejack 16-May-14 8:09am

How does it make sense?

CPallini 16-May-14 8:13am

It doesn't make sense (to me) to define the template specializations within the template class declaration itself.

Stefan_Lang 16-May-14 8:23am

Asking why a form of template definition make sense is a bit like asking a fish why it doesn't swim backwards. ;-)

Seriously though: you've got to remember that the code inside your template class only gets analyzed by the compiler if it stumbles upon an instantiation of that class, and in that case it creates a class definition using the actual template arguments. I. e., at the time it looks at the member function, the class template arguments already are determined! Therefore, if you want a template class member function to behave differently for a specific instantiation, you need to define this outside the class definition.

Sergejack 16-May-14 10:30am

Ok, I got it. I had no real clue how the internal mechanics worked.

Stefan_Lang 16-May-14 8:17am

My 5. I wasn't even aware you could specialize an individual function rather than the entire class. Then again, i guess it wouldn't mak much sense otherwise...