CODEWITHSUNDEEP
c++functionclassobject
TPRammus
TPRammus

Reputation: 511

Can you pass a function so you can later call it?

I want to have objects with one method which calls a function (but every object should have a different function to call). I will try to show you what I mean by showing an example:

class Human
{
    public:
        void setMyFunction(void func);  // specify which function to call
        void callMyFunction();  // Call the specified function
};

void Human::setMyFunction(void func)    // ''
{
    myFunction = func;
}

void Human::callMyFunction()    // ''
{
    myFunction();
}

void someRandomFunction()   // A random function
{
    // Some random code
}

int main()
{
    Human Lisa;     // Create Object
    Lisa.setMyFunction();   // Set the function for that object
    Lisa.callMyFunction();  // Call the function specified earlier
}

This code (obviously) doesn't work but I hope you understand what I am trying to accomplish.

MfG, TPRammus

Upvotes: 3

Views: 91

Answers (2)

Jarod42
Jarod42

Reputation: 217275

You might use std::function.

#include <functional>

class Human
{
    std::function<void()> mFunc;
public:
    void setMyFunction(std::function<void()> func) { mFunc = func; }
    void callMyFunction() { if (mFunc) mFunc(); }
};

Demo

Upvotes: 5

SergeyA
SergeyA

Reputation: 62583

I would suggest using a simple function pointer. Just do this:

class Human
{
    public:
        using func_t = void (*)(); 
        void setMyFunction(func_t f) {
             func = f;
        }
        void callMyFunction() {
             func();
        }
     private:
        func_t func;
};

The reasons why one might prefer function pointers to std::function are:

  • Performance. Calling std::function tends to be slower, than calling a function by pointer.
  • std::function needs truly ugly syntax when one needs to bind it to an overloaded function.

Example:

void foo();
void foo(int x = 0);

void check() {
  Human h;
  h.setMyFunction(&foo);
}

Will fail to compile.

Upvotes: 4

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