how to use functors as ordinary functions argument?

[laikon]

dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

Thanks a lot.

[red floyd]

laikon wrote:

dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

overload f.

#include <functional>
void f(int a, int b, const std::binary_function<int, int, int>& f)
{
std::cout << f(a,b) << std::endl;
}

[Paul Brettschneider]

red floyd wrote:

laikon wrote:

>dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

overload f.

#include <functional>
void f(int a, int b, const std::binary_function<int, int, int>& f)
{
std::cout << f(a,b) << std::endl;
}

Or, since the function body is the same, use templates?

#include <iostream>

template <typename Tvoid f(int a, int b, const T &fp)
{
std::cout << fp(a, b) << std::endl;
}

int mult(int a, int b)
{
return a * b;
}

int main()
{
int x = 2, y = 3;

f(x, y, std::plus<int>());
f(x, y, mult);
}

[Gianni Mariani]

laikon wrote:

dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

Turn f into a template ?

[Juha Nieminen]

laikon wrote:

dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

Make it a template:

template<typename Functor>
void f(int a, int b, Functor fp)
{
std::cout << fp(a, b) << std::endl;
}

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2008-04-01 19:00, laikon wrote:

dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

Use a function, things that uses functors are built using templates.

--
Erik Wikström

[Pete Becker]

On 2008-04-01 13:50:03 -0400, red floyd <no*****@here.dudesaid:

>
overload f.

#include <functional>
void f(int a, int b, const std::binary_function<int, int, int>& f)
{
std::cout << f(a,b) << std::endl;
}

That won't work. std::binary_function has three typedefs, nothing more.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[red floyd]

Pete Becker wrote:

On 2008-04-01 13:50:03 -0400, red floyd <no*****@here.dudesaid:

>>
overload f.

#include <functional>
void f(int a, int b, const std::binary_function<int, int, int>& f)
{
std::cout << f(a,b) << std::endl;
}

That won't work. std::binary_function has three typedefs, nothing more.

Oops... you're right. My bad.

[red floyd]

Juha Nieminen wrote:

laikon wrote:

>dear, all:

below is a function with a parameter of function pointer.

void f(int a, int b, int (*fp)(int, int))
{
std::cout << fp(a, b) << std::endl;
}
and f is called with functor plus as 3rd argument.

int x = 2, y = 3;
f(x, y, std::plus<int>());

but it raises an error, and how to solve it?

Make it a template:

template<typename Functor>
void f(int a, int b, Functor fp)
{
std::cout << fp(a, b) << std::endl;
}

Actually, since the OP wanted an int return:

template<typename Functor>
void f(int a, int b, Functor fp)
{
std::cout << static_cast<int>(fp(a,b)) << std::endl;
}

[Juha Nieminen]

Kai-Uwe Bux wrote:

I guess, one could devise some template magic to implement and check the
conceptual requirement that Functor has an operator( int, int ) that
returns an int. However, I think that would be overkill in most cases.

If I'm not mistaken, this will become easy with the next C++ standard.