Menu driven program

I am trying out a menu driven program which has following options:
1.Factorial of a number
2.Prime or not
3.Odd or even
4.Exit
Once a menu item is selected the appropriate action should be taken and once the action is finished ,the menu should reappear.Unless the user selects the exit option the program should continue to work.
I have typed the following program but the menu item does not reappear again.

Code:

---

#include<stdio.h>
#include<stdlib.h>
void main(exitmenu)
{
int opt,p,j,x=1,num,i=2,eo;
choice:
while(x<4)
{
printf("\n1.Factorial of a number");
printf("\n2.Prime number or not");
printf("\n3.Odd or Even");
printf("\n4.Exit");
printf("\nEnter your choice");
scanf("&#37;d",&opt);

switch(opt)
case 1:
{
 p=1,j=1;
 while(j<=7)
 {
        p=p*j++;
        printf("\nFactorial of %d=%d",x++,p);
 }
 break;
case 2:
{
 printf("\nInput a number");
 scanf("%d",&num);
 while(i<num && num%i!=0)
 i++;
 if(num==i)
 printf("Prime number");
 else
 printf("Not a prime number");
 break;
}
case 3:
{
printf("\nInput a number");
scanf("%d",&eo);
if(eo%2==0)
printf("\nNumber is even");
else
printf("\nNumber is odd");
break;

}



case 4:
exit(exitmenu);
goto choice;
}
}
}

---

What will be the right one??

> void main(exitmenu)
This is all wrong.
main returns an int, and the parameters are either
int main ( int argc, char *argv )
int main ( void )

Rather than trying to cram everything into a single function, try something like

Code:

---

opt = getMenuChoice();  // prints the menu, reads input, returns result
switch ( opt ) {
  case 1:
    doFactorial();
    break;
  case 2:
    doPrime();
    break;
  case 3:
    printf( "TODO - Odd or Even\n" );
    break;
}

---

If you start with all your cases as being a printf, then you can check that all your menu stuff works first without cluttering up the code.
When it does work, you can then confidently move onto each sub-function and test each one of those in turn.

You're being bitten by scanf() leaving the newline in the stdin input buffer.

Todd

What is the meaning of:
dofactorial();

Just eyeballing this code, option #2 won't work a second time through the code. Variable i will have a residual value from the first time through.

Quote:

---

Originally Posted by Todd Burch

You're being bitten by scanf() leaving the newline in the stdin input buffer.

Todd

---

Then what should I do?

Quote:

---

Originally Posted by paushali

Then what should I do?

---

Read the forum's FAQ on the best way to get input from the user.

Todd

And it does for all cases except 1.

Your case-label is on the wrong side of the brace, by the way - not that it really makes any difference, but it looks neater if it's on the right side.

Your indentation is horrible (as in, doesn't exist).

Why do you have a exit followed by a goto? The goto will never be reached if exit() works, right?

The return type of main is int - see the FAQ.

--
Mats

Quote:

---

Originally Posted by matsp

Why do you have a exit followed by a goto? The goto will never be reached if exit() works, right?

--
Mats

---

OK I have deleted that goto but still its not working.

And which (if any) of the other 6 replies did you follow the advice of?

Did you look at your case 1: code to see if you could figure out why it's exiting ONLY when you choose that case, and not 2 or 3? [I mean, I didn't even write the code, and I spotted the error after half a minute or so]

--
Mats

> What is the meaning of:
> dofactorial();
It means it is a function you should write to implement that part of your assignment.
You can of course call it whatever you want.