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Base conversion in C

  1. Dec 27, 2012 #1
    I have been working on coding a program that works for all bases, including non-integers. I have been checking it against wolframalpha, and so far, my code works for a decent amount of bases, but at the same time has trouble with others... For instance it seems to have a tolerance in the magnitude and the digit length of the bases. So, far I have tested and it seems to work for bases larger than 1.5 (I have tested it up to 64 so far) and up to 6 digit bases(6 including the digits after the decimal) with some exceptions. I have noticed that for the decimal bases that my program computes that don't match wolfram, the string will match until a certain digits place, say k, and at k, k-1 wolfram will have a 10 respectively, while my computed value will have the floor of the base in the k-1 place. I figured that it may be that my algorithm wasn't finding the highest power divisor right, but then I realized that would not make since. So, I was hoping that fresh eyes may be able to help me out with it... By the way, this is just a rough draft, but the procedure should be fairly simple to follow.

    Code (Text):


    #include <stdio.h>
    #include <math.h>

    void convert_base(double x, double b)
    {
        double y, z;
        int digit, i, j = 0, k;

    /* Find the terms in power series expansion with base power > 0 */

        z = x;
        while(z >= 1)
        {

    /* Compute largest power of base which divides x  giving a value >= b*/

            y = z;
            digit = y;
            i = 0;
            while(digit >= b)
            {
                y = y / b;
                digit = y;
                i++;
                //printf("y = %lf\n", y);
            }

            //printf("i = %d\n", i);
            //printf("digit = %d\n", digit);

    /* Subtract off this term in series expansion */

            z = z - (digit * pow(b, i));

            //printf("z = %lf\n", z);

    /* Fill in digits in between the last calculated digit and this calculated digit */

            if((abs(i - j) > 1) && (j != 0))
            {
                for(k = 1; k < abs(i - j); k++)
                {
                    printf("0");
                }
            }
            printf("%d", digit);

            j = i;

        }

        if(j != 0)
        {
            for(k = j; j > 0; j--)
            {
                printf("0");
            }
        }

        printf(".");

    /* Find the terms in the power series expansion with base power < 0 */

        while((z > 0) && (z <= 1) && (i > -6))
        {

    /* Compute smallest power of base which multiplies x giving a value >1 */

            y = z;
            digit = y;
            i = 0;
            while(digit < 1)
            {
                y = y * b;
                digit = y;
                i--;
                //printf("y = %lf\n", y);
            }

            //printf("i = %d\n", i);
            //printf("digit = %d\n", digit);

    /* Subtract off this term in series expansion */

            z = z - (digit * pow(b, i));

            //printf("z = %lf\n", z);

    /* Fill in digits in between the last calculated digit and this calculated digit */

            if(abs(i - j) > 1)
            {
                for(k = 1; k < abs(i - j); k++)
                {
                    printf("0");
                }
            }
            printf("%d", digit);

            j = i;
        }

        if(z == 0)
        {
            printf("000");
        }

        printf("...\n");
    }

    int main(void)
    {
        double x, b;

        printf("Enter a number: x = ");
        scanf("%lf", &x);
        printf("Enter the base for conversion: b = ");
        scanf("%lf", &b);

        printf("(x)_b = %lf = ", x);

        convert_base(x, b);

        return 0;
    }

     
     
  2. jcsd
  3. Dec 28, 2012 #2
    Code (Text):
    if(j != 0)
        {
            for(k = j; j > 0; j--)
            {
                printf("0");
            }
        }
    sorry, I'm not that familiar with the subtleties of C ... why have you got j>0;j-- in what appears to be the k loop?

    Can you give a couple of examples of where your code differs from WolframAlpha?
     
  4. Dec 28, 2012 #3
    Sorry about that, I was trying to format the code in the physics forum box and add comments and I guess I messed that loop up. But here are some examples:

    10 in base 3.14159:
    My code: [tex](x)_{b} = (10.000000)_{3.14159} = 30.121201...[/tex]

    Wolfram: [tex]100.01022123001..._{3.14159} [/tex]


    1230293 in base 3.10293

    My code: [tex](x)_{b} = (1230293.000000)_{3.10293} = 1120103012012.122122...[/tex]

    Wolfram: [tex]1.1201100020201..._{3.10293}×3.10293^{12}[/tex]

    The last one really shows what I was talking about the digits are the same until the 3 in mine which you notice is the floor of the base while theirs is a 1 in the preceding digits place and 0 where I have a 3, I pointed the floor of the base thing out because all we are really doing is modular arithmetic. And I am guessing it has something to do with that. You will notice that after that mine differs from theirs completely. In the first one this happens at the first digit, so the entire string is different.
     
    Last edited: Dec 28, 2012
  5. Dec 28, 2012 #4
    I was thinking that it may actually be machine error.
     
  6. Dec 28, 2012 #5
    I figured that I would post a few where they are the same also:

    3623847 in base 39:

    My code: [tex](x)_{b} = (3623847.000000)_{39} = 1223216.000...[/tex]

    Wolfram: [tex]1223216_{39}[/tex]

    10 in base 4.16273

    My code: [tex](x)_{b} = (10.000000)_{4.16273} = 21.231202...[/tex]

    Wolfram: [tex]21.2312021013..._{4.16273}[/tex]

    10 in base 2.718281828:

    My code: [tex](x)_{b} = (10.000000)_{2.718282} = 102.1120101...[/tex]

    Wolfram: [tex]102.1120101111..._{2.71828}[/tex]
     
    Last edited: Dec 28, 2012
  7. Dec 28, 2012 #6

    I like Serena

    User Avatar
    Homework Helper

    They are both proper representations of the numbers involved.
    The difference is that Wolfram tries to get the highest possible digits in the highest possible places and apparently your algorithm does not.

    Note that if the base is a whole number, the representation is unique.
    But if the base is a fractional number, there is more than 1 representation.

    You can verify this by multiplying out the numbers.
     
  8. Dec 28, 2012 #7
    Oh thank you! I did not know that. Is there somewhere that I can read more about this?
     
  9. Dec 28, 2012 #8

    I like Serena

    User Avatar
    Homework Helper

    Sorry, I don't really know.
    I've learned about it in old books that you won't have available.
    So what's left is googling for it.
     
  10. Dec 28, 2012 #9
    I'll do that when I have some free time. Thanks for the information though. It's always a good feeling when you get a program right the first time through :)
     
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