printf problem

This is a discussion on printf problem within the C Programming forums, part of the General Programming Boards category; A friend ask me to help with his code. the input will be as following: 1st input -> number of ...

  1. #1
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    Apr 2005
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    printf problem

    A friend ask me to help with his code. the input will be as following:

    1st input -> number of loop
    2nd input -> given number
    3rd input -> number of rotation

    example:

    1
    123045
    3

    The result is 045123

    He told me that he can only get 45123 as the result. the 0 is missing. Is there any way to display it ?? i'm not very familiar with C or C++ though

    here is his code. thx in advance

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int main()
    {
    long int num,n,value, digitNum, ctr, NumX;
    int left, right;
    long b;
    
    scanf("%ld", &b);
    
    while (b != 0)
    {
          
    scanf("%d %d", &num, &n);
    
    value = num;
    NumX = 1;
    
         while (value!= 0)
         {
          NumX = NumX * 10;
          value = value / 10;
         }
         
         NumX = NumX / 10;
         
         ctr = 0;
         
         while (ctr != n)
         {
               left = num / 10;
               right = num % 10;
               num = right * NumX + left;
               ctr++;
               
               }
               
               printf("%d", num);
               
    printf("\n");
    b--;
    }
    
    fflush(stdin);getch();
    return 0;
    }
    Last edited by blackmisery; 11-01-2006 at 02:01 PM.

  2. #2
    CSharpener vart's Avatar
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    Maybe it is easyer to convert number to string and work with the string?

    As for the original question - yes using format modifiers the number can be printed with zeroes filling it up to desired length
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  3. #3
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    can u explain more about this format modifiers ??? is it something like

    Code:
    {
        ("%.2f", variable) ;
    }
    i'm not very sure as well.

    and for the code, my friend told me that his so called "Genius" lecture insist of using number instead of string. This because my friend is only in his 2nd month of his new semester.

    i tried to help him by using array but still this "genius" lecture reject it. he still insist to use number without array or anything.

  4. #4
    CSharpener vart's Avatar
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    yes it is format modifier
    http://msdn.microsoft.com/library/de..._functions.asp

    read about width and preceiding zeroes
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  5. #5
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    Code:
    {
        ("%.2f", variable) ;
    }
    as far as i knw abt this if the statement is like as follows

    Code:
     printf  ("%.2f", variable) ;
    the value after the float will be ristricted to two digits

    the answer wud be sumthing like
    10.22

  6. #6
    Registered User SKeane's Avatar
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    What about something like ...

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    
    int main()
    {
        unsigned int i, num, num_digits, num_loops, num_rots;
        int t;
    
        printf("Enter number of loops : ");
    
        if ( scanf("%u", &num_loops) != 1 )
        {
            /* Error */
        }
    
        for (i = 0; i < num_loops; i++)
        {
            printf("Enter number to rotate : ");
    
            if ( scanf("%u", &num) != 1 )
            {
                /* Error */
            }
    
            if ( num == 0 ) num_digits = 1;
            else            num_digits = 1 + (int) log10( (double) num );
    
            printf("Enter number of rotations : ");
            if ( scanf("%u", &num_rots) != 1 )
            {
                /* Error */
            }
    
            printf("Rotated number is : ");
            for (i = num_digits; i > 0; i--)
            {
                t = i + num_rots - 1;
                if ( t >= num_digits) t -= num_digits;
                printf("%u", (num / (int)pow(10,t)) % 10);
            }
            printf("\n");
        }
    
        getch();
    
        return 0;
    }

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