Thread: need help debugging a program counting bits set from an integer number

  1. #1
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    need help debugging a program counting bits set from an integer number

    The program is supposed to count the number of bits set by an integer. for example: 5 is 00000101 and has two bits set. compiling the program with debug statements makes me think that the program isn't even entering the integer into the loop

    Code:
    /********************************************************
     *							*
     * FILE: ch10_ex4.c					*
     * CREATED: 8/15/07					*
     * BY: bpf						*
     *							*
     *  this program counts the number of bits set in an 	*
     *   integer						*
     *							*
     ********************************************************/
    
    #include<stdio.h>
    
    main()
    {
      char line[100];
      int itgr; /* the integer value of the input number */
    
      int r; /* remainder */
      int set = 0; /* number of variables set */
      int n; /* index into the line of bits */
      int subt = 0; /* the value to subtract from the remainder starts at zero each loop */
    
      /* input number */
      (void)printf("enter an integer: ");
      (void)fgets(line, sizeof(line), stdin);
      (void)sscanf(line, "%d", &itgr);
    
      r = itgr; /* remainder starts off the same as the integer */
    
      while(0)
        {      
          /* if the remainder is equal to one, the one's column is set and the loop can stop */
          if (r == 1)
    	{
    	  set++;
    
    	  break;
    	}
    
          /* if the remainder is equal to zero the number of bits set are counted and the loop stops */
          else{ if (r == 0)
    	  {
    	    break;
    	  }
    
    	/* the highest multiple of two is counted up to and becomes the number subtracted from the remainder */      
    	else{ for (n = 1; n <= r; n * 2)
    	    {
    	      subt = n;
    	    }
    
    	  /* the remainder is lowered and one column is set */	  
    	  r = r - subt;
    	  set++;
    	}
          }
        }
    
      /* how many bits are set */
      (void)printf("%d bits are set\n", set);
    
      return(0);
    }

  2. #2
    and the hat of int overfl Salem's Avatar
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    How many times do you think while(0) will execute ?
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  3. #3
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    until the remainder goes to 1 or zero

    so the for loop will find the highest multiple of two and subtract it from the remainder until it reaches the 'one's' spot

  4. #4
    Kernel hacker
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    What exactly is the condition inside the parenthesis of "while(0)", and what does that evaluate to as a "true/false" statement?

    --
    Mats

  5. #5
    Chinese pâté foxman's Avatar
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    What basically people are trying you to answer is that 0 means "false" in C and everything else means "true". So, by writing while(0), you are basically saying "while false" which, of course, evalute to false...

    And by the way, this is an infinite loop if r > 1
    Code:
    for (n = 1; n <= r; n * 2)
    {
        subt = n;
    }
    instead, you should write something like
    Code:
    for (n = 1; n <= r; n = n * 2)
    {
        subt = n;
    }
    Also, you could rewrite your code so it have a nicer and easier to understand structure/logic. But it's not that bad either...

    Also, you might want to check the return value of sscanf and check if the number entered by the user is non negative before entering your loop.

  6. #6
    Registered User ssharish2005's Avatar
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    Code:
     while(itgr)
      {
          if(itgr&#37;2 == 1)
             set++;
          
          itgr /= 2;   
      }
    You could replace your while loop with this one. This minimizes your code. And why do want to type cast the printf return value to void. There is no need for that.

    ssharish2005

  7. #7
    Registered User
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    lol while(0) is ironic. (you'd think he wanted to use while since he put it in there, but apparently not.)

    it's like thinking "i'm going to open the door" while standing in front of the door doing nothing.

    ...lol


    while(1) is ... probably more effective.
    Last edited by simpleid; 08-17-2007 at 03:24 PM.

  8. #8
    Cogito Ergo Sum
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    you can just use // to comment out, saves time and fingers moving around to do /* and */ everytime u want to comment

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