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Recursive function trace

Posted on 2007-11-16
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Last Modified: 2010-04-15
Hey guys,

I have little experience with recursion and was wondering if someone could help me figure out how to trace it. I need to obtain its output by calling it with a particular parameter, such as 7, for example. Is this as simple as putting printf() statements or am I missing something big?


int MyFunc(int x) {
 

	if (x <= 1) {

		return (1);

	} else {

		return (x * MyFunc(x - 2));	

	}

}

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Question by:Koraxen
10 Comments
 
LVL 10

Expert Comment

by:peetm
ID: 20298833
Something like this?

#include<stdio.h>

int MyFunc(int x)
{
    printf("In MyFunc x = %d\n", x);

        if (x <= 1)
        {
            return 1;
        }
        else
        {
                int d = MyFunc(x - 2);
               
                printf("Tail from 'else', x = %d d = %d\n", x, d);
               
                return x * d;
        }
}

int main(void)
{
    printf("%d\n", MyFunc(10));
   
    getchar();
   
    return 0;
}
0
 
LVL 53

Assisted Solution

by:Infinity08
Infinity08 earned 200 total points
ID: 20298872
Yes, you can just add printf statements to show the progress. Take care to place them in the correct location though. More specifically : if you place the printf somewhere before the recursive call, it will be shown in the order that the functions are recursively called - if you place the printf somewhere after the recursive call, it will be shown in the order that the recursive calls are unwound.

The next code for example will show for input 7 :

7
5
3
1
int MyFunc(int x) {

        printf("%d\n", x);

        if (x <= 1) {

                return (1);

        } else {

                return (x * MyFunc(x - 2));     

        }

}

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Author Comment

by:Koraxen
ID: 20301210
I understand. I think this might be a trick question though, because it seems too straightforward. Given that snippet of code, the actual question is:

What is the value of A, where "A = SpecialFunction (7)"

I can easily throw this into the compiler and run it, and get 105.
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Expert Comment

by:Infinity08
ID: 20301257
>> I can easily throw this into the compiler and run it, and get 105.

Yes. And what's your question ?
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Author Comment

by:Koraxen
ID: 20302464
It just sounds like one of those questions that's too easy to be true. Either I'm expected to trace it by hand or there's some trick to it that I'm not seeing.
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Expert Comment

by:Infinity08
ID: 20302477
Is this an assignment ? If so, can you post the exact wording of it ?
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Author Comment

by:Koraxen
ID: 20302501
Yes.

=============
Consider the following function:

int MyFunc(int x) {
      if (x <= 1) {
            return (1);
      } else {
            return (x * MyFunc(x - 2));      
      }
}

What is the value of A where "A = MyFunc(7)" ?
=============
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Expert Comment

by:Infinity08
ID: 20302556
Then you've already got the answer - you posted it earlier : 105.

If you need to add a printf to the code, then just do this :

        printf("%d", MyFunc(7));
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Expert Comment

by:billtouch
ID: 20323922
Just for fun, you might add a counter and increment it upon entry and decrement it on exit. Include the counter in your printf(), and you can watch your recursion in action.

Working from peetm's version:

int MyFunc(int x)
{
    int ret;
    static counter=0;
    counter++;
    printf("(%d) In MyFunc x = %d\n", counter, x);

    if (x <= 1) {
        ret = 1;
    } else {
        int d = MyFunc(x - 2);
        printf("(%d) Tail from 'else', x = %d d = %d\n", counter, x, d);
        ret = (x * d);
    }
    counter--;
    return ret;
}

int main(void)
{
    printf("%d\n", MyFunc(10));
    getchar();
    return 0;
}


Bill
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Accepted Solution

by:
itsmeandnobodyelse earned 300 total points
ID: 20327668
>>>> It just sounds like one of those questions that's too easy to be true.

MyFunc(7) = 7 * MyFunc(5) =
                    7 * 5 *MyFunc(3) =
                    7 * 5 * 3 * MyFunc(1) =
                    7*5*3*1 =
                   105.

Do you have any problems with that?

Regards, Alex

     
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