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Integer array length


 
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# 8  
Old 09-09-2013
For the same reason. The two numbers displayed on each call of p_array() stem from the fact that a[] is an 8-byte pointer and sizeof(int) is 4 bytes, and 8/4 = 2.

So, it's displaying the first two elements of the array as you're sorting it in qsort().

Note that you passed len to qsort(). Why can't you pass it to q_array()?
# 9  
Old 09-09-2013
Thanks Scott!
Yes, I did pass len to p_array, but did not make any difference. Here my focus is about the calculation of the array length.
I could not understand why the calculation of array length in the main function is fine (line 10), but not in the p_array function (line 66). Thanks!
# 10  
Old 09-09-2013
In the main() function it's an array. In the p_array() function it's a pointer to the first element of the array.

I didn't see where you passed len to p_array(), or how it was used there, so I can't comment on it.
# 11  
Old 09-09-2013
Sorry I meant I tried passing len to the p_array function like this:
Code:
#include<stdio.h>
void qsort(int[], int, int);
void p_array(int[], int);

int main()
{
    int i;
    int array[] = { 55, 66, 32, 12, 9, 73, 2, 4, 37, 36 };
    int len = sizeof(array) / sizeof(int);
    for (i = 0; i < len; i++)
        printf("%d ", array[i]);
    printf("\n");
    p_array(array, len);
return 0;
}

void p_array(int arr[], int size)
{
    int i;
    size = sizeof(arr) / sizeof(int);
//    int size = sizeof (arr) / sizeof (arr[0]);
    for (i = 0; i < size; i++)
        printf("%d ", arr[i]);
    printf("\n");
}

output:
Code:
55 66 32 12 9 73 2 4 37 36 
55 66

I did not catch the part you said it is an array in the main() but pointer in the p_array(). Seems that's the part I missed. Thanks a lot!
# 12  
Old 09-09-2013
You still have this, which will override the passed variable:
Code:
    size = sizeof(arr) / sizeof(int);

Also, as you originally passed len - 1 to qsort(), you'll need to modify the for-loop in p_array to:
Code:
    for (i = 0; i <= size; i++)

This User Gave Thanks to Scott For This Post:
# 13  
Old 09-09-2013
Thanks again! That bug only make small difference. Output:
Code:
55 66 32 12 9 73 2 4 37 36 
55 66 32

I feel understand pointer and array in C (not fluent to use, though!). Could you explain why the two calculations differ?
Code:
int len = sizeof(array) / sizeof(int);  //10
int  size = sizeof(arr) / sizeof(int);   //2

Thanks.

Last edited by yifangt; 09-09-2013 at 01:09 PM..
# 14  
Old 09-09-2013
I'm not sure where you get 8 from - you'll have to show the code (in context) that produces that number, but passing the length works perfectly fine:
Code:
# cat main.c
#include <stdio.h>

int main( int argc, char **argv ) {
  int array[10] = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 };
  int i, len = sizeof( array ) / sizeof( int );
  printf( "In main, length of array: %d\n", len );
  printf( "In main, array elements:" );
  for( i = 0; i < len; i++ )
    printf( " %d", array[i] );

  printf( "\n" );
  printf( "\n" );

  showarr_1( array );
  showarr_2( array, 10 );

  return 0;
}
 
int showarr_1( int array[] ) {
  int len = sizeof( array ) / sizeof( int );
  printf( "In showarr_1, length of array: %d (bogus number!)\n", len );
  printf( "\n" );

  return 0;
}
 
int showarr_2( int array[], int len ) {
  int i;
  printf( "In showarr_2, length of array: %d\n", len );
  printf( "In showarr_2, array elements:" );
  for( i = 0; i < len; i++ )
    printf( " %d", array[i] );

  printf( "\n" );

  return 0;
}


# ./main
In main, length of array: 10
In main, array elements: 3 6 9 12 15 18 21 24 27 30

In showarr_1, length of array: 2 (bogus number!)

In showarr_2, length of array: 10
In showarr_2, array elements: 3 6 9 12 15 18 21 24 27 30

edit: you updated your post, to change 8 to 10. It should be clear now why the numbers are different. As previously said: one is an array, the other is a pointer (to the array).
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