Thread: [Newbie] Factorial

Hi, I'm doing my homework and I'm stuck on this factorial part. I know this is the code for n! from my text:

Code:

#include <stdio.h>
double  mfact(int);
main()
{
           int m;
           printf("Enter an integer m>0: ");
           scanf("%d",&m);
           printf("m! is: %g\n",mfact(m));
}
double mfact(int m)
{
    if(m==1)return 1;
    else return m*mfact(m-1);
}

But I need 2n! instead so I tried to modify the code to this:

Code:

#include <stdio.h>
double  mfact(int);
main()
{
           int m;
           printf("Enter an integer m>0: ");
           scanf("%d",&m);
           printf("m! is: %g\n",mfact(m));
}
double mfact(int m)
{
    if(m==1)return 2;
    else return 2m*mfact(2m-1);
}

Because I figured that 2n! = 2n*(2n-1) ... etc. so it would work but it's messed up. So in fact, I realized I don't get what this whole code is about. What is the mfact specifically?

Thanks for the help. I don't know why but I'm still getting the wrong answer because when I entered in 2, I get 48 when I'm supposed to get 24.

Code:

#include <stdio.h>
double  mfact(int);
main()
{
           int m, n;
           printf("Enter an integer m>0: ");
           scanf("&#37;d",&n);
	     m = 2*n;
           printf("2n! is: %g\n",mfact(m));
}
double mfact(int m)
{
if(m==2)return 2;
else return m*mfact(m-1);
}

Thanks! I got it now!

Edit: Just a quick question but is it possible to put m = 2n; like this:

Code:

double mfact(int m)
{
m = 2n;
if(m==2)return 2;
else return m*mfact(m-1);
}

You should just write mfact() to compute the factorial of any integer >= 0. Then just plug in 2*n when you call it. You're not going to accomplish or simplify anything by trying to change the body of the function itself. BTW, if you replaced "if (m == 1)" with "if (m == 0)" you could have it handle the case 0! == 1.