Help with a recursive function that computes a to the n

This is a discussion on Help with a recursive function that computes a to the n within the C Programming forums, part of the General Programming Boards category; Hello! I have been working on this for several days now and the program has no errors but it does ...

  1. #1
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    Help with a recursive function that computes a to the n

    Hello!

    I have been working on this for several days now and the program has no errors but it does not calculate. Here's the code:
    Code:
    #include "stdafx.h"
    #include "stdio.h"
     
    float power(float a, int n); // function prototype//
     
     
    int main (void)
    {
        float a;
     
        int n;
     
        printf("\n enter a value of a");
    	scanf ("%f", &a);
    	printf("\n enter value of n");
    	scanf ("%f", &n);
        printf("\n %.4f to the power of % d is %.4f",a,n,power(a,n)); 
    	
    	 return 0;
    }
    float  power(float a, int n)
    {
    
        if (a==0) 
        {
            return 0;
     
        }
        else if(n==0)
        {
            return 1;
     
        }
        else if (n>0)
        {
            return( a* power(a,n-1));
        }
        else
        {
            return ((1/a)*power(a,n+1));
        }
    }
    Please help me figure this out!
    Last edited by Salem; 10-02-2008 at 11:38 AM. Reason: Added [code][/code] tags, learn how to use them yourself!!!!!

  2. #2
    Registered User C_ntua's Avatar
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    What's seems to be the problem? What is the output?

  3. #3
    Banned master5001's Avatar
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    >> return ((1/a)*power(a,n+1));

    Really?

    The other problem with doing it this way is that your order of operations is out of whack too.
    Last edited by master5001; 10-02-2008 at 11:48 AM.

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    Quote Originally Posted by master5001 View Post
    >> return ((1/a)*power(a,n+1));

    Really?

    The other problem with doing it this way is that your order of operations is out of whack too.
    No that looks right. a^-3 = (1/a)*a^-2 = (1/a)*(1/a)*a^-1=(1/a)*(1/a)*(1/a)*1.

    He could compute 1/(a^n) in that final condition, but this should work almost as well.
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

  5. #5
    Banned master5001's Avatar
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    Ah ok, fair enough, it is going "less than 0." Though I am still going to point out that one could theoretically be compounding floating point errors. I would still only do my division once to minimize floating point errors. But I am also one of those crazy types of guys who would typically subtract 1.0f from a floating value too....

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    Yes, that's why I said almost as well. But if the concern is precision, then the first thing to fix is to change float to double.
    It is too clear and so it is hard to see.
    A dunce once searched for fire with a lighted lantern.
    Had he known what fire was,
    He could have cooked his rice much sooner.

  7. #7
    Banned master5001's Avatar
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    Either way, I am not spotting anything that would cause any sort of unusual returns. I would use a double for this. And even with a double I would still do: return 1.0 / power(a, -n); But that is just me.

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    When you're inputting n, being an integer, you should use

    scanf ("%d", &n);
    instead of
    scanf ("%f", &n);

    [sound of "doh!!"]

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    The issue is when you run the program.

    Quote Originally Posted by C_ntua View Post
    What's seems to be the problem? What is the output?
    The problem is that the program does not seem to have the correct output. I attached a screenshot to explain what I mean!

    Thanks!
    Attached Images Attached Images  

  10. #10
    Banned master5001's Avatar
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    Quote Originally Posted by nonoob View Post
    When you're inputting n, being an integer, you should use

    scanf ("%d", &n);
    instead of
    scanf ("%f", &n);

    [sound of "doh!!"]

    Uh oh... Todd has some competition for the title of Ol' Eagle Eye. That looks like reason enough for your code to be broken by my count.

    Being in a good mood today, here is a recap of all issues we have addressed.

    Suggestions implemented:
    Code:
    #include "stdafx.h"
    #include "stdio.h"
     
    float power(float a, int n); // function prototype//
     
     
    int main (void)
    {
        float a; 
        int n;
     
        printf("\n enter a value of a");
        scanf ("%f", &a);
        printf("\n enter value of n");
        scanf ("%d", &n);
        printf("\n %.4f to the power of % d is %.4f",a,n,power(a,n)); 
    
        return 0;
    }
    float  power(float a, int n)
    {
        /* I put the numbers in float notation. Not a big thing. I am nit-picking. */
        if (a==0.0f) 
        {
            return 0.0f;
     
        }
        else if(n==0)
        {
            return 1.0f;
         }
        else if (n>0)
        {
            return( a* power(a,n-1));
        }
        else
        {
            return (1.0f / power(a,-n)); // 1 extra recursion. 1 division
        }
    }

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    Thank you all for your help! Now, what can I do to turn this into a non-recursive function.

  12. #12
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    Write it in a non-recursive way. If only there was a way to repeat a certain group of code while some condition remains true....

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    Ooooh ooooh ooooh pick me pick me!!!

    Gotos?

  14. #14
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by nonoob View Post
    Ooooh ooooh ooooh pick me pick me!!!

    Gotos?
    Now, now, Johnny, you know what happens when you say ... that word.

    /me throws tennis ball at head.

    Anyone else?

  15. #15
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    Here is the code I am trying to compute for a non- recursive function now!
    #include "stdafx.h"
    #include "stdio.h"

    double mypower(float a, int n); // function prototype//


    int main (void)
    {
    float a;
    int n;
    int power;

    printf("\n enter a value of a\n");
    scanf ("%d", &a);
    printf("\n enter value of n\n");
    scanf ("%d", &n);
    power=mypower (a,n);
    printf("The result is %d\n", power);

    return 0;
    }
    double power(float a, int n)
    {
    if (a==0.0)
    {
    return 0.0;

    }
    else if(n==0)
    {
    return 1.0;
    }
    else if (n>0)
    {
    return( a* mypower(a,n-1));
    }
    else
    {
    return (1.0/ mypower(a,-n));
    }
    }


    The error is error LNK2019: unresolved external symbol "double __cdecl mypower(float,int)" (?mypower@@YANMH@Z) referenced in function _main
    C:\Documents and Settings\Owner\My Documents\Visual Studio 2008\Projects\lab 8.3.2\Debug\lab 8.3.2.exe : fatal error LNK1120: 1 unresolved externals

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