Recursive factorial return statement

Question

Why do we use return 1 to terminate the recursive function? Can any other value be used as a default value like 1.

And if we return 1 as return value of the function, then why 1 is not returned to main function.

 #include<stdio.h>
 int fact(int n)
 {
   if(n>=1)
      return (n*fact(n-1));
   else
      return 1;
 }
 int main()
 {
   int a,ans;
   scanf("%d",&a);
   ans=fact(a);
   printf("factorial of %d is %d ",a,ans);
   return 0;
  }
  /*
   explanation
          fact(4);
          if(4>=1) 4*fact(3)
          if(3>=1) 4*3*fact(2)
          if(2>=1) 4*3*2*fact(1)
          if(1>=1) 4*3*2*1*fact(0)
          if(0>=1) return 1;

  */

Your comment is the answer. You are calling fact(4), not fact(0) — mch
– mch, Commented May 21, 2019 at 8:30
please only tag the language you are using. Your question is rather unclear. Try fact(0) and use a debugger to see that 1 is indeed returned to main — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented May 21, 2019 at 8:30

axiac · Accepted Answer · 2019-05-21 08:49:45Z

int fact(int n)
{
    if (n >= 1)
        return n * fact(n-1);
    else
        return 1;
}

Each time the fact() function is called, it runs either the return n * fact(n-1); statement OR the return 1; statement but not both.

You call fact(4) in main(). This is how it runs:

main:
  compute fact(4)
  fact(4):
  |  4 >= 1?  Yes!
  |  compute 4 * fact(3)
  |    fact(3):
  |    |  3 >= 1?  Yes!
  |    |  compute 3 * fact(2)
  |    |    fact(2):
  |    |    |  2 >= 1? Yes!
  |    |    |  compute 2 * fact(1)
  |    |    |    fact(1):
  |    |    |    |  1 >= 1? Yes!
  |    |    |    |  compute 1 * fact(0)
  |    |    |    |    fact(0):
  |    |    |    |    |  0 >= 1? NO!
  |    |    |    |    |  return 1;
  |    |    |    |    +--> 1
  |    |    |    |  fact(0) is 1, return 1 * 1 (--> 1)
  |    |    |    +--> 1
  |    |    |  fact(1) is 1, return 2 * 1 (--> 2)
  |    |    +--> 2
  |    |  fact(2) is 2, return 3 * 2 (--> 6)
  |    +--> 6
  |  fact(5) is 6, return 4 * 6 (--> 24)
  +--> 24
  fact(4) is 24, assign it to `ans`, print it etc
// end of main

When a function uses the return statement (or after it executes its last statement if a return is not reached, the control is passed back to the expression that has called it.

Blaze · Accepted Answer · 2019-05-21 08:35:28Z

3

why do we use "return 1" to terminate the recursive function

Because this is supposed to cover the case for when n is not >=1, in other words when n is 0. I don't think negative n are valid. 0! is 1, hence why it returns that value.

And if we return 1 as end of function, then why 1 is not returned to main function.

If the function is called with 0 or 1 for n, then 1 is returned to the main function. For any other value, 1 is only returned in the recursive factorial calls and the value which is returned to the main function isn't 1, but (n*fact(n-1)), which isn't 1 in those cases.

edited May 21, 2019 at 8:35

answered May 21, 2019 at 8:31

Blaze

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2 Comments

Sourav Ghosh Over a year ago

then why 1 is not returned to main function... I think that's a bit different from what you explained in the last paragraph.

Blaze Over a year ago

Good catch. I'll fix it.

Rishikesh Raje · Accepted Answer · 2019-05-21 08:30:34Z

2

The return statement is executed when n==0.

The factorial for n==0 is 1, so we return 1.

answered May 21, 2019 at 8:30

Rishikesh Raje

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Comments

selvabharathi s · Accepted Answer · 2019-05-21 08:43:48Z

   /*
      explanation
          fact(4);
          if(4>=1) 4*fact(3)
          if(3>=1) 4*3*fact(2)
          if(2>=1) 4*3*2*fact(1)
          if(1>=1) 4*3*2*1*fact(0)
          if(0>=1) return 1; now return is default tend to multiply as we give 1 and 
           return has already 24 in its stack so 1*24 is returned to main()
          if we give return 2; 2*24 is returned to main();

    */

we don't want our final result to be affected, to solve that error, anything multiple by 1 will result the same, so we use 1 as return in recursive function.

Actually return is also a stack register which holds temporary variable while calling with in the function, and it operates default by multiply property, as we can send only one return value always.

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