converting to base 10

This is a discussion on converting to base 10 within the C Programming forums, part of the General Programming Boards category; Hi, I'm currently writing a program that takes a number (n) and a base (b) between [1, 10] from the ...

  1. #1
    xp5
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    converting to base 10

    Hi,

    I'm currently writing a program that takes a number (n) and a base (b) between [1, 10] from the user's input and convert it into base 10. However, I have to make sure that all the digits of n has to be between [0, b-1] with the exception of n being 1. For example:

    n = 012 and b =3, output is 5
    n = 00110 and b=2, output is 6
    n = 00110 and b=1, output is 2

    for base (b) = 1, the current program cannot check to see if the digits of 'n' is greater than 1 or not because of this code that i use to check it: output[index] = n%b. For example:

    n=23 and b = 2 : output[index] = 23%2 = 3 ; and 3 is greater than base 2
    BUT if n = 23 and b =1; ouput[index] = 23%1 = 0; and I CANNOT TAKE THE 3 to compare to base 1 so that 3 is greater than 1.

    Please help me! The following is my code:

    Code:
    #include<stdio.h>
    
    int main(void)
    {
      int n, b;
      int output[64];
      int index = 0;
    
      printf("Enter base(b) between [1,10], and a number(n) between [0,b-1] in this format 'n b': ");
      scanf("%d %d", &n, &b);
    
      printf("n is: %d\n", n);   //Test
      printf("b is: %d\n", b);   //TEST
    
      if ((b < 1) || (b > 10))
      {
        printf("Your base is not between 1 and 10");
        return 0;
      }
    
      if (n < 0)
      {
        printf("n must be positive");
        return 0;
      }
    
    
      while (n != 0)
      {
        output[index] = n % b;
    
        if ((output[index] >= b) && (output[index] != 1))  //TROUBLE AREA RIGHT HERE
        {
          printf("one of the digits is greater than the base");
          return 0;
        }
    
        n = n/b;
        ++index;
        printf("number to convert: %d\n", n);
      }
      return 0;
    }
    Thank you!

  2. #2
    and the hat of wrongness Salem's Avatar
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    Well the if() inside the while loop makes no sense, since it can never be true. You've just done n&#37;b, so output[index] >= b will always be false.

    Also, what is the other printf() supposed to do?
    Printing output[index] would tell you more IMO.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  3. #3
    xp5
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    Sorry, the while loop is supposed to look like this:
    Code:
    #include<stdio.h>
    
    int main(void)
    {
      int n, b;
      int output[64];
      int index = 0;
    
      printf("Enter base(b) between [1,10], and a number(n) between [0,b-1] in this format 'n b': ");
      scanf("&#37;d %d", &n, &b);
    
      printf("n is: %d\n", n);   //Test
      printf("b is: %d\n", b);   //TEST
    
      if ((b < 1) || (b > 10))
      {
        printf("Your base is not between 1 and 10");
        return 0;
      }
    
      if (n < 0)
      {
        printf("n must be positive");
        return 0;
      }
    
    
      while (n != 0)
      {
        output[index] = n % 10;
    
        if ((output[index] >= b) && (output[index] != 1))  //TROUBLE AREA RIGHT HERE
        {
          printf("one of the digits is greater than the base");
          return 0;
        }
    
        n = n/b;
        ++index;
        printf("number to convert: %d\n", n);
      }
      return 0;
    }
    For some reason if i put n = 110 and b=2, it goes into the if statement, which it is not supposed to.

  4. #4
    Captain Crash brewbuck's Avatar
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    This problem is WAY easier if you read the "n" value as a string instead of an integer. You are going to be processing it digit-wise, anyway, so why convert from digits to a value only to convert back to digits again?

  5. #5
    xp5
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    Hi, I'm reading "n" as a string now, but for some reason I'm keep getting a "segmentation fault (core dumped)". Do you know why is it doing this?

    Code:
    #include<stdio.h>
    
    int main(void)
    {
      int b;
      char n;
      int output[64];
      int index = 0;
    
      printf("Enter base(b) between [1,10], and a number(n), such that digits of n is between[0,b-1] in this format 'n b': ");
      scanf("&#37;s %d", &n, &b);
    
      printf("n is: %s\n", n);   //Test
      printf("b is: %d\n", b);   //TEST
    /*
      if ((b < 1) || (b > 10))
      {
        printf("Your base is not between 1 and 10");
        return 0;
      }
    
      if (n < 0)
      {
        printf("n must be positive");
        return 0;
      }
    
    
      while (n != 0)
      {
        output[index] = n % 10;
        n = n/b;
        ++index;
        printf("number to convert: %d\n", n);
      }*/
      return 0;
    }
    Thank you!

  6. #6
    Kernel hacker
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    You're not reading it as a string, you're reading it as a single char (but with string format).

    --
    Mats

  7. #7
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by xp5 View Post
    Hi, I'm reading "n" as a string now, but for some reason I'm keep getting a "segmentation fault (core dumped)". Do you know why is it doing this?
    n needs to be an array large enough to hold the input. E.g:

    Code:
    char n[64];
    Also, you'll have to change other parts of the code, because processing digit-by-digit is completely different than your old method. But the result will be much simpler than what you have now.

    Since I suspect this is homework, I won't post an actual solution -- but you should be able to solve the entire problem in fewer than 20 lines of code. The function I wrote which converts the number string from its given base is only three lines long.

  8. #8
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    Quote Originally Posted by brewbuck View Post
    n needs to be an array large enough to hold the input. E.g:

    Code:
    char n[64];
    Also, you'll have to change other parts of the code, because processing digit-by-digit is completely different than your old method. But the result will be much simpler than what you have now.

    Since I suspect this is homework, I won't post an actual solution -- but you should be able to solve the entire problem in fewer than 20 lines of code. The function I wrote which converts the number string from its given base is only three lines long.
    There's even a library function to do that, right?

    By the way, is base 1 actually a valid base? What's the semantics of "base 1"?

    --
    Mats

  9. #9
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by matsp View Post
    By the way, is base 1 actually a valid base? What's the semantics of "base 1"?
    Tick marks.

    1 = 1
    2 = 11
    3 = 111
    4 = 1111
    5 = 11111
    etc...

    EDIT: Base 1 was probably the first method of written counting ever invented. Even more amusing is the concept of "Base 0," where each number gets its own unique symbol:

    1 = Harry
    2 = Sam
    3 = Bobby
    4 = Johnny
    5 = Terrence
    etc...
    Last edited by brewbuck; 08-30-2007 at 04:58 PM.

  10. #10
    Registered User ssharish2005's Avatar
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    For some reason if i put n = 110 and b=2, it goes into the if statement, which it is not supposed to.
    Thats becuase of this

    Code:
    n = n/b;  /* Which should have been n/10
    But any way you should ry this problem with the solution, using string. What other sugguested you.

    HINT: The thing which u are trying to do can be done using strtol function.
    ssharish2005

  11. #11
    xp5
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    Actually, I can't use char in this problem, everything has to be integer. But, I got it now! Thanks!!

  12. #12
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    Quote Originally Posted by xp5 View Post
    Actually, I can't use char in this problem, everything has to be integer. But, I got it now! Thanks!!
    Consider that you may want to input base > 10, in which case the normal numbers don't suffice, so you you need to read the original number in as a string - not a single char of course, but an array of chars (which, when it's terminated with a zero(0 non '0') is a string in C).

    --
    Mats

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