How Can I Pass Function as an Argument? C/C++

It's a small syntax issue.  :)

  exitApplication(&mainMenu());

That tries to run the function mainMenu and return the address of "something".  Probably not what you want to do.

  exitApplication(mainMenu);

That passes the address of mainMenu to exitApplication.  That's probably what you want to do.

Note that if mainMenu is a class method, it can get somewhat more complicated.


Good Luck,
Kent

Change "exitApplication(&mainMenu());" by "exitApplication(mainMenu);"

I did it, i got this error:

error C3867: 'MenusLayout::mainMenu': function call missing argument list; use '&MenusLayout::mainMenu' to create a pointer to member

mainMenu resides in MenusLayout class anyway, so why should i do call the class as MenusLayout::?

Here is the code as you see it below...

This a simple command line remote administration tool for my project.

Everytime user want to exit, he/she will be asked for confirmation. If he/she presses no, user should in the same page. The only to do it, is to call the same page (function) that user is curently in.

#include <sstream>
#include "MenusLayout.h"
#include "ConsoleDesign.h"
 
const int EXIT = -1;
ConsoleDesign windowDes;
 
MenusLayout::MenusLayout()
{
	//empty constructor
}
 
MenusLayout::~MenusLayout()
{
	//empty destructor
}
 
 
void MenusLayout::mainMenu()
{
	do
	{
		choice_is_invalid = false;
 
		windowDes.coloringText(3);
		cout<<"\nEnter a choice: ";
		windowDes.coloringText(7);
 
		int receiveUserInput = getUserChoice<int>();
 
		switch(receiveUserInput)
		{
			case 1:
				//.......
				break;
			case 2:
				//.......
				break;
			case 3:
				//.......
				break;
			case EXIT:
                                     //*********HERE
				exitApplication(mainMenu);
				break;
			default:
				windowDes.coloringText(6);
				cout<<"Choice is invalid."<<endl;
				choice_is_invalid = true;
				break;
		}
	}while(choice_is_invalid);
 
}
 
void MenusLayout::buildRemoteController()
{
	do
	{
		choice_is_invalid = false;
 
		windowDes.coloringText(3);
		cout<<"\nEnter a choice: ";
		windowDes.coloringText(7);
 
		int receiveUserInput = getUserChoice<int>();
 
		switch(receiveUserInput)
		{
			case 1:
				//.......
				break;
			case 2:
				//.......
				break;
			case 3:
				system("cls");
				windowDes.headerDesign();
				mainMenuDisplay();
				mainMenu();
				break;
			case EXIT:
                                     //******HERE
				exitApplication(buildServer); 				break;
			default:
				windowDes.coloringText(6);
				cout<<"Choice is invalid."<<endl;
				choice_is_invalid = true;
				break;
		}
	}while(choice_is_invalid);
 
}
 
 
//*************HERE
void MenusLayout::exitApplication(void (*app)(void))
{
	do
	{
		choice_is_invalid = false;
 
		windowDes.coloringText(12);
		cout<<"\nAre you sure you want to exit - Y/N? ";
 
		char exitConfirmation = getUserChoice<char>();
 
		switch(exitConfirmation)
		{
			case 'y':
				exit(0);
				break;
			case 'Y':
				exit(0);
				break;
			case 'n':
				app();
				break;
			case 'N':
				app();
				break;
			default:
				windowDes.coloringText(6);
				cout<<"Choice is invalid - Please enter Y or N"<<endl;
				choice_is_invalid = true;
				break;
		}
	}while(choice_is_invalid);
}

Select allOpen in new window

There's a difference between a function pointer, and a member function pointer.

So :

void MenusLayout::exitApplication(void (MenusLayout::*app)(void)) {
    (*this.*app)();
}
 
MenusLayout ml;
ml.exitApplication(&MenusLayout::mainMenu);

Select allOpen in new window

Thanks jkr & Infinity08. My code worked. Since this is first time implementing  such a code, i have some questions...

Thats how i understand it:

void MenusLayout::exitApplication(void (MenusLayout::*app)(void))

this means the parameter will point to a function. Thus, the passed argument be be preceded with & (to pass the address).

When user passes the address of the function, this will be executed (this->*app)();

1. (this->*app)(); means the function resides in the scope (MenusLayout) and will be pointing to where function resides - Can you explain this part please i do not understand it fully.

2. exitApplication(&MenusLayout::mainMenu);

Function address is passed. Why we have to use MenusLayout? mainMenu resides inside MenusLayout. I think because MenusLayout has its own address, while its member function have their own addresses. Right?

3. But, why this wouldn't work exitApplication(&mainMenu);? Pointer eventually will point to the target function we not need to go through MenusLayout address first. Can you explain please?

4. Assume, we have MenusLayout::mainMenu() { //implementation}
MenusLayout::designApplication() { //implementation}

Usually if i want to call mainMenu in designApplication, i just call it normally as:

MenusLayout::designApplication() { mainMenu(); } ....

Why it does not force me to call it as MenusLayout::designApplication() { MenusLayout::mainMenu(); }?

Thanks for being patience

Can i call you a life saver?