C++ using function as parameter

Question

Possible Duplicate:
How do you pass a function as a parameter in C?

Suppose I have a function called

void funct2(int a) {

}


void funct(int a, (void)(*funct2)(int a)) {

 ;


}

what is the proper way to call this function? What do I need to setup to get it to work?

You have two functions, which one are you having problems calling? Note that you have hidden the function `funct2` in `funct` by having an identically named pointer-to-function parameter so you will have to fully qualify `funct2` to call it directly from inside `funct`. — CB Bailey, Jun 14 '11 at 06:50
This question might help: http://stackoverflow.com/questions/9410/how-do-you-pass-a-function-as-a-parameter-in-c — Jim, Jun 14 '11 at 06:46
Right on, because C++ is C as we all know. Welcome to CloseOverflow. In case one uses C++11 there is good read here: http://stackoverflow.com/questions/16111285/how-to-pass-and-execute-anonymous-function-as-parameter-in-c11 -- avoiding passing functions with pointers improves readability. — greenoldman, Jun 03 '14 at 12:44
A more complete set of answers: http://stackoverflow.com/questions/16111285/how-to-pass-and-execute-anonymous-function-as-parameter-in-c11 — Eponymous, Mar 09 '17 at 18:39
How is this a duplicate of the referenced question about C? It is possible that C++ provides (or at some point will provide) a different approach. — pooya13, Jan 14 '19 at 21:01
@pooya13 It's not a duplicate. There are other ways to pass functions as parameters in C++ that are not possible in C (using [functors](https://www.quantstart.com/articles/Function-Objects-Functors-in-C-Part-1), for example). — Anderson Green, Nov 22 '19 at 17:32

Matt Razza · Accepted Answer · 2011-06-14T06:55:53.683

37

Normally, for readability's sake, you use a typedef to define the custom type like so:

typedef void (* vFunctionCall)(int args);

when defining this typedef you want the returning argument type for the function prototypes you'll be pointing to, to lead the typedef identifier (in this case the void type) and the prototype arguments to follow it (in this case "int args").

When using this typedef as an argument for another function, you would define your function like so (this typedef can be used almost exactly like any other object type):

void funct(int a, vFunctionCall funct2) { ... }

and then used like a normal function, like so:

funct2(a);

So an entire code example would look like this:

typedef void (* vFunctionCall)(int args);

void funct(int a, vFunctionCall funct2)
{
   funct2(a);
}

void otherFunct(int a)
{
   printf("%i", a);
}

int main()
{
   funct(2, (vFunctionCall)otherFunct);
   return 0;
}

and would print out:

edited Jun 14 '11 at 06:55

answered Jun 14 '11 at 06:49

Matt Razza

3,524
2
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11

Since C++11, one would rather use `using vFunctionCall = void (*)(int args);` for better readability. Or even better, use [std::function](https://stackoverflow.com/questions/25848690/should-i-use-stdfunction-or-a-function-pointer-in-c) – Frederico Pantuzza Sep 15 '17 at 08:29
4

There is no need to do any casting. – HAL9000 Sep 20 '20 at 19:15
1

Since C++11 I would rather like to see std::function<> – Goswin von Brederlow Oct 04 '20 at 10:28
2

@GoswinvonBrederlow `std::function` does *much more* than `void (*)(int args)`, and is therefor harder for the platform to optimise – Caleth Nov 23 '20 at 12:37

score 32 · Answer 2 · edited Aug 03 '23 at 10:37

32

Another way to do it is using the functional library.

std::function<output (input)>

Here reads an example, where we would use funct2 inside funct:

#include <functional>
#include <iostream>

void displayMessage(int a) {
    std::cout << "Hello, your number is: " << a << '\n';
}

void printNumber(int a, std::function<void (int)> func) {
    func(a);
}

int main() {
    printNumber(3, displayMessage);
    return 0;
}

output :

Hello, your number is: 3

edited Aug 03 '23 at 10:37

Toby Speight

27,591
48
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103

answered Sep 20 '20 at 19:04

Marine Galantin

1,634
1
17
28

1

Perhaps you should demonstrate calling the function – JVE999 Oct 29 '20 at 00:11
3

Try not to use `using namespace std`. It's bad practice: https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice?rq=1 – Overt_Agent Jul 06 '21 at 13:15
Can I use this syntax to pass a public method of class A to the constructor of class B and if yes, how? – tickietackie Mar 28 '22 at 19:36

score 0 · Answer 3 · edited Aug 20 '23 at 19:58

0

A more elaborate and general (not abstract) example

#include <iostream>
#include <functional>
int Add(int a, int b)
{
    return a + b;
}
int Mul(int a, int b)
{
    return a * b;
}
int Power(int a, int b)
{
    while (b > 1)
    {
        a = a * a;
        b -= 1;
    }
    return a;
}
int Calculator(std::function<int(int, int)> foo, int a, int b)
{
    return foo(a, b);
}
int main()
{
    cout << std::endl
         << "Sum: " << Calculator(Add, 1, 2);
    cout << std::endl
         << "Mul: " << Calculator(Mul, 1, 2);
    cout << std::endl
         << "Power: " << Calculator(Power, 5, 2);
    return 0;
}

edited Aug 20 '23 at 19:58

Michał Leon

2,108
1
15
15

answered Aug 31 '22 at 10:28

Pawan Kumar

114
9

I feel like this doesn't add enough value compared to [this answer](https://stackoverflow.com/a/63982640/11107541) to warrant being a separate answer. – starball Sep 02 '22 at 18:57

score -3 · Answer 4 · answered Jun 14 '11 at 06:50

-3

check this

typedef void (*funct2)(int a);

void f(int a)
{
    print("some ...\n");
}

void dummy(int a, funct2 a)
{
     a(1);
}

void someOtherMehtod
{
    callback a = f;
    dummy(a)
}

answered Jun 14 '11 at 06:50

Tatvamasi

2,537
19
14

That's C, not C++ – Michał Leon Aug 20 '23 at 19:08

C++ using function as parameter

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