Thread: Newbie function help please

I am currently learing C with cprogramming.com's "C made easy" tutorial. I am trying to call a function in a switch case statement and when I try to compile it gives me "too few arguments to function" error. Can someone please help me figure this out?

Thanks in advance!

Code:

#include <stdio.h>


int playgame ( int input );
int loadgame ( int input );
int playmultiplayergame ( int input );


int main()
{
int input;

printf("1. Play game\n");
printf("2. Load game\n");
printf("3. Play Multiplayer\n" );
printf("4. Exit Game\n");
printf("Selection: ");
scanf("%d", &input);
switch ( input ) {
	case 1:
		 playgame( input );
		break;
	case 2:
		loadgame( input );
		break;
	case 3:
		playmultiplayergame( input );
		break;
	case 4:
		printf("Thanks for playing\n");
		break;
	default:
		printf("Bad input, quitting!\n");
		break;
		
	}
int playgame ( int input )
	{
	printf("You selected play game. Have fun!\n");
	}
	
	int loadgame ( int input )
	{
	printf("Load game from where?\n");
	}
	
	int playmultiplayergame ( int input )
	{
	printf("Who would like to play with?\n");
	}

}

Code:

*/ Working code */
#include <stdio.h>


int playgame ( int input );
int loadgame ( int input );
int playmultiplayergame ( int input );


int main()
{
int input;

printf("1. Play game\n");
printf("2. Load game\n");
printf("3. Play Multiplayer\n" );
printf("4. Exit Game\n");
printf("Selection: ");
scanf("%d", &input);
switch ( input ) {
	case 1:
		 playgame( input );
		break;
	case 2:
		loadgame( input );
		break;
	case 3:
		playmultiplayergame( input );
		break;
	case 4:
		printf("Thanks for playing\n");
		break;
	default:
		printf("Bad input, quitting!\n");
		break;
		
	}
} */ To here */

int playgame ( int input )
{
printf("You selected play game. Have fun!\n");
}
	
int loadgame ( int input )
{
printf("Load game from where?\n");
}
	
int playmultiplayergame ( int input )
{
printf("Who would like to play with?\n");
}

*/ } moved from here */

You have placed the main() end tag "}" in the end of the program. You have to
have it where the main function ends.

> Can someone please help me figure this out?
Yeah, really big

Originally Posted by Laars

Code:

*/ Working code */

Oh no it's not. Not as of the first character on the first line it isn't.


Quzah.

shouldn't he also have the return 0; at the end to?

and im not sure if you know or not, but you can also put your functions before mian.

Code:

#include <stdio.h>

int playgame (int input)
{
    printf("You selected play game. Have fun!\n");
}
	
int loadgame (int input)
{
    printf("Load game from where?\n");
}
	
int playmultiplayergame (int input)
{
    printf("Who would like to play with?\n");
}

int main()
{
    int input;

    printf("1. Play game\n");
    printf("2. Load game\n");
    printf("3. Play Multiplayer\n" );
    printf("4. Exit Game\n");
    printf("Selection: ");
    scanf("%d", &input);

    switch (input)
    {
        case 1:
	    playgame(input);
	    break;
	case 2:
	    loadgame(input);
	    break;
	case 3:
            playmultiplayergame(input);
	    break;
	case 4:
	    printf("Thanks for playing\n");
	    break;
	default:
            printf("Bad input, quitting!\n");
	    break;
    }

    return (0);
}

Edit: opps i was just gonna make evertying else with a short but i forgot to...

> short input;
Should be an int, given the usage everywhere else in the code (especially the scanf which is now broken).

On all functions that return int but main(), you need a return 0. If there is no return statement in main(), return 0 is assumed. That doesn't cover void main(), though.

[edit]
That doesn't cover your other functions that return int, either. Make them void or return something.
[/edit]

You must never have void main()

If you want a portable program on a hosted implementation, yea. But never say never, because you can get bitten by an exception.

return 0 is not assumed, unless they've changed this for C99.

C99 does make this change.