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How do you create a menu that uses an array of function pointers?

Posted on 2008-10-30
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Last Modified: 2013-12-14
I'm trying to create a program with a menu that uses an array of function pointers.  So when the user selects an option the appropriate function is called.  I know that I am close to figuring this out, but just haven't quite got it.  (I know that I could use a switch statement but I'm suposed to use this array.)  Please tell me what's wrong with my code.  Also is there another way that I can initialize each element in my char array to "none"?
const int size = 10;
const int length = 25;
char Name[size][length] = {"none","none","none","none","none","none","none","none","none","none"};
 
int main()
{
	int Option = 0;
		
	int (*fptr[6])(char[size][length]) = {Add, Update, Delete, Sort, Print, Quit};//declaration creating array of fuction pointers
	
	do 
	{ 
		system("cls");
		cout << "Main Menu",
		cout << "\n",
		cout << "1) Add\n",
		cout << "2) Update\n",
		cout << "3) Delete\n",
		cout << "4) Sort\n",
		cout << "5) Print\n",
		cout << "6) Quit\n" << endl;
		
		cout << "Enter Selection" << endl;
		cin >> Option;
 
		if(Option < 1 || Option > 6)
		{
			cout << "invalid entry - please re-enter" << endl;
		}		else
		{
			cout << fptr[Option - 1]() << endl;
		}
 
 
		}while(Option != 6);
 
	return 0;
 
}

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Question by:b_acs
6 Comments
 
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Expert Comment

by:mgonullu
ID: 22838911
Why you are doing in the hard way and not just use Switch statement?
Regarding the initialization of your array why not use a nested for loop
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Expert Comment

by:Zoppo
ID: 22838932
Hi b_acs,

I think you need to write '(*fptr[Option - 1])()' to call the functions via their pointer.

Hope that helps,

ZOPPO
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Accepted Solution

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Infinity08 earned 200 total points
ID: 22838933
When you call the function through the function pointer, you need to provide the arguments. In this case you need to pass a 2D char array :

                        cout << fptr[Option - 1](Name) << endl;




>> Also is there another way that I can initialize each element in my char array to "none"?

Instead of putting "none" in there, I'd initialize the whole array to '\0' bytes, so they are all empty strings. If you want to show "none", you can always check whether the string is empty, and show "none" in that case :

        if (*name) {
            /* show name */
        }
        else {
            /* show "else" */
        }
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Expert Comment

by:Infinity08
ID: 22838937
>> Why you are doing in the hard way and not just use Switch statement?

Using function pointers for this kind of stuff is cleaner. You don't need a switch.
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Author Closing Comment

by:b_acs
ID: 31511568
I knew I was close!  Thank you!
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Expert Comment

by:Infinity08
ID: 22839036
Btw, instead of having something like this :

        int (*fptr[6])(char[size][length])

do yourself a favor, and make your life a little easier by using typedef's :

        typedef int (*MenuFunction)(char[size][length]);

This typedef is the function pointer as you defined it. You can then simply create an array of function pointers like this :

        MenuFunction fptr[6] = { ... };

Much easier to read and understand ;)
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