Recursion ( No return still works)

Asked Nov 23 '16 at 19:00

Active Nov 23 '16 at 19:00

Viewed 48 times

I was trying to write a function which will take an integer and return it's factorial. I used c++ language. The actual code should be:

    int f(int x)
{
    if(x==0) 
        return 1;
     else if(x>=1)    
        return x=x*f(x-1);
    else
        return 1;
}

This is correct form. But,

int f(int x)
{
   if(x>1)
      x=x*f(x-1);
}

this also works even I didn't mentioned any 'return'. Also when the recursion goes to base case, there is nothing to return! So, how does it works actually ? :o

asked Nov 23 '16 at 19:00

Shohidul Islam Bulbul

It doesn't work. It won't compile. – nicomp Nov 23 '16 at 19:02
@nicomp, nah, it may compile. – SergeyA Nov 23 '16 at 19:03
What does 'work' mean for you? – SergeyA Nov 23 '16 at 19:03
2

@nicomp It _may compile_ (with a warning), but is undefined behavior. – πάντα ῥεῖ Nov 23 '16 at 19:03
It invokes undefined behavior. WIth that you were very *unlucky* it seems to work (because it doesn't). – WhozCraig Nov 23 '16 at 19:03
@WhozCraig, there is no way in the world it might 'work' with any sane meaning of word 'work;. – SergeyA Nov 23 '16 at 19:04
@Sam I have to add that dupe to my list. Great finding! – πάντα ῥεῖ Nov 23 '16 at 19:04
If it work, then you're just lucky. Launching the program a second time could simply crash. Or it could format your hard drive, or spit out baby elephants from your computer case. The compiler is LITERALLY permitted to do anything. – Guillaume Racicot Nov 23 '16 at 19:06
@SergeyA that would be why I said "seems to work" and that "it doesn't". And if that isn't obtuse enough, I honestly couldn't care less. – WhozCraig Nov 23 '16 at 19:06
This has answers in the way local variables are stored in memory. It is undefined but I think many of you may find this simple example works. Google where 'return' values are taken from. – kabanus Nov 23 '16 at 19:10

Recursion ( No return still works)

0 Answers0