1. We have moved from vBulletin to XenForo and you are viewing the site in the middle of the move. Though the functional aspect of everything is working fine, we are still working on other changes including the new design on Xenforo.
    Dismiss Notice

strange program- shows not output

Discussion in 'C' started by k0der, Jul 13, 2010.

  1. k0der

    k0der New Member

    Joined:
    Apr 14, 2009
    Messages:
    19
    Likes Received:
    0
    Trophy Points:
    0
    Hi,

    the following code compiles well but its nt showing output. Please specify what may be the error..


    Code:
    #include<stdio.h>
    
      #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
      int array[] = {23,34,12,17,204,99,16};
    
      int main()
      {
          int d;
    
          for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
              printf("%d\n",array[d+1]);
    
          return 0;
      }
    Thanks
     
  2. Poonamol

    Poonamol New Member

    Joined:
    Mar 31, 2010
    Messages:
    33
    Likes Received:
    0
    Trophy Points:
    0
    In the for loop, value of D must be initialized to 0 instead of -1.
    for(d=0;........)
    :happy:
     
  3. k0der

    k0der New Member

    Joined:
    Apr 14, 2009
    Messages:
    19
    Likes Received:
    0
    Trophy Points:
    0
    thats fine.. but what is going wrong when d=-1 ?

    Thanks
     
  4. virxen

    virxen New Member

    Joined:
    Nov 24, 2009
    Messages:
    387
    Likes Received:
    90
    Trophy Points:
    0
    try this

    Code:
    #include<stdio.h>
    
      #define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
      int array[] = {23,34,12,17,204,99,16};
    
      int main()
      {
          int d;
          int t=(TOTAL_ELEMENTS-2);
          printf("\ntotal elements=%d\n",t);
          for(d=-1;d <= t;d++)
              printf("%d\n",array[d+1]);
          return 0;
      }
    
    
     
  5. xpi0t0s

    xpi0t0s Mentor

    Joined:
    Aug 6, 2004
    Messages:
    3,012
    Likes Received:
    203
    Trophy Points:
    0
    Occupation:
    Senior Support Engineer
    Location:
    England
    It goes wrong because TOTAL_ELEMENTS evaluates to an unsigned int. This also displays no output:
    Code:
    	int d;
    
    	unsigned int t=TOTAL_ELEMENTS-2;
    	for(d=-1;d <= t;d++)
    		printf("%d\n",array[d+1]);
    
    You compiler should give you a warning. I got (in Visual Studio 2008):
    warning C4018: '<=' : signed/unsigned mismatch

    The problem with comparing signed with unsigned is that the signed number is treated as unsigned. -1 is 0xFFFFFFFF, which as an unsigned int has the value over 4 billion, which is not less than or equal to TOTAL_ELEMENTS (7), so you get no output.

    No, this is absolutely false. It makes no difference what the start and stop values of d are, as long as a suitable offset is added when d is used as the index of an array, IF the pointer used is at the start of the array. This works perfectly:
    Code:
    	int d, *arr=&array[3];
    	
    	for (d=-3; d<4; d++)
    		printf("%d ",arr[d]);
    	printf("\n");
    	for (d=-3; d<4; d++)
    		printf("%d ",array[d+3]);
    
    Both loops are from -3 to +3 (7 elements). The first loop displays arr[-3] to arr[3], which corresponds to array[0] to array[6] because arr points at array[3]. The second loop displays array[-3+3] to array[3+3], i.e. array[0] to array[6].
     
  6. k0der

    k0der New Member

    Joined:
    Apr 14, 2009
    Messages:
    19
    Likes Received:
    0
    Trophy Points:
    0
    Thanks a ton xpi0t0s!! Its was a great info!!:happy:
     
  7. k0der

    k0der New Member

    Joined:
    Apr 14, 2009
    Messages:
    19
    Likes Received:
    0
    Trophy Points:
    0
    Code:
     int main()
      {
          int* p;
          p = (int*)malloc(sizeof(int));
          *p = 10;
          return 0;
      }
    Though this works fine on IA-32, but does not work on IA-64. I am not able to find the answer,why it is so.

    Please tell me.

    Thanks
     
  8. xpi0t0s

    xpi0t0s Mentor

    Joined:
    Aug 6, 2004
    Messages:
    3,012
    Likes Received:
    203
    Trophy Points:
    0
    Occupation:
    Senior Support Engineer
    Location:
    England
    How exactly doesn't it work? There is no output. What are you expecting, and what did you observe that didn't match your expectations? What OS are you using - IA32/IA6 is a processor architecture, and runs several different OS's.
     
  9. jimblumberg

    jimblumberg New Member

    Joined:
    May 30, 2010
    Messages:
    120
    Likes Received:
    29
    Trophy Points:
    0
    Could it be that sizeof(int) != sizeof(int *) in 64 bit?
     
  10. xpi0t0s

    xpi0t0s Mentor

    Joined:
    Aug 6, 2004
    Messages:
    3,012
    Likes Received:
    203
    Trophy Points:
    0
    Occupation:
    Senior Support Engineer
    Location:
    England
    Why would that make any difference? p is a pointer to int. malloc allocates space for an int and returns the pointer, cast to an int* to avoid compiler warnings, into p. The next line then dereferences p and plonks 10 into it. Should work fine, i.e. set *p, which is an int, to 10.

    In fact the code should be identical to
    Code:
    int p;
    p=10;
    malloc(sizeof(int)); // to duplicate the memory leak in the original code
    
     
  11. k0der

    k0der New Member

    Joined:
    Apr 14, 2009
    Messages:
    19
    Likes Received:
    0
    Trophy Points:
    0
    it was asked by someone to me. I have only IA32.

    The int* is unsigned integer (as pointing to an address) and int both have same size of 64bits on IA-64. So at least i do not see any problem with the code.
     
  12. xpi0t0s

    xpi0t0s Mentor

    Joined:
    Aug 6, 2004
    Messages:
    3,012
    Likes Received:
    203
    Trophy Points:
    0
    Occupation:
    Senior Support Engineer
    Location:
    England
    Perhaps that's the problem. An int* is NOT an unsigned integer; it is a pointer to a signed integer. These are not the same thing. So maybe the problem is with the interpretation of code (int*=unsigned int) rather than the code itself.

    On a 64-bit platform pointers may be 64-bit. int may still only be 32-bit - it depends on how the compiler author implements the 32 to 64-bit upgrade. Upgrading ints to 64-bit might (a) be unnecessary and (b) break too much existing code, so for example Microsoft have chosen to leave int as 32-bit and you need a different type if you actually want a 64-bit integer.
     
    Last edited: Jul 19, 2010

Share This Page