please help sort array won't work

Hey There. I'm pretty new to C and I was trying to write this sort program. It will display the first line nicely, but it won't display lines pass 1 through pass 7. Can anyone see any errors with anything? should I have the call in the for loop.
Please help.

Code:

---

#include <stdio.h>

#define SIZE 8

void swap(int*,int*);

void sort(int a[], int n);


int main(void)
{
      int i,cnt=0,a[SIZE] = {7,3,66,3,-5,22,-77,2};

      printf("Unsorted Data:");
      for(i=0; i<SIZE; i++){
            printf("  %i", a[i]);
      }
      putchar('\n');
      sort(a,SIZE);
      for(i=0; i<SIZE; i++){
          ++cnt;
            printf("After Pass:  %d", cnt);
            printf("  %i", a[i]);
      }
      putchar('\n');
      return 0;
}


void sort(int a[], int n)
{

      int i,j;

      for(i=0; i<SIZE; ++i)
          for (j = i + 1; j<SIZE; ++j)
        if (a[i] > a[j])
            swap(&a[i], &a[j]);
       
       
}

void swap(int *p, int *q)
{
      int tmp;
       
      tmp = *q;
      *p = *q;
      *q = tmp;
}

---

You need braces around those for loops, there are multiple statements that you want to loop.

around the ones in the sort function or main(void). I'm confused because after it prints out the unsorted data (all the numbers of the array), it will only print out the first number of the array for the remainder of the passes.

No, no braces are missing. They just need to indent so you can see it:

Code:

---

void sort(int a[], int n)
{
    int i,j;

    for(i=0; i<SIZE; ++i) /* This one... */
        for (j = i + 1; j<SIZE; ++j) /* grabs this, which... */
            if (a[i] > a[j]) /* grabs this, which... */
                swap(&a[i], &a[j]); /* grabs this. */
}

---

Quzah.

am I printing out the results of each pass through the sort incorrectly in the main(void)?

I suppose that depends on what you mean by "incorrectly". Just copy and paste your above print loop. You aren't printing each pass of the sort. You're printing the sorted version once. Look:

Code:

---

print the array
sort array
print the array

---

That's what you're doing. Your sort function only runs once, and you're printing the array after you sort it. Not while you are sorting.


Quzah.

This is what the output looks like as I have it now:
(I moved the putchar('\n') inside the braces)

output:

Code:

---

Unsorted Data: 7 3 66 3 -5 22 -77 2
After pass: 1    -77
After pass: 2    -77
After pass: 3    -77
After pass: 4    -77
After pass: 5    -77
After pass: 6    -77
After pass: 7    -77
After pass: 8    2

---

I don't understand why it won't print out all the elements of the array.?

I just cant seem to figure out a way to print the array as it makes each pass through the sort function. Does anyone have any ideas?

It doesn't matter where you move the \n. You're still only displaying it after you call sort. If you want it to print while it sorts, add printing into your sort function.


Quzah.

Like quzah says, you need to put a printf() loop inside the sort function, for example:

Code:

---

#include <stdio.h>


void sort (int a[], int n);

int main (void)
{
  int a[] = {7, 3, 66, 3, -5, 22, -77, 2};

  sort (a, sizeof a / sizeof *a);
  return 0;
}


void sort (int a[], int n)
{
  int i, j, k, tmp;

  for (i = 0; i < n; ++i)
  {
    printf ("Pass %d:", i);
    for (k = 0; k < n; ++k)
      printf ("% 5d", a[k]);
    putchar ('\n');

    for (j = i + 1; j < n; ++j)
      if (a[i] > a[j])
      {
        tmp = a[i];  a[i] = a[j];  a[j] = tmp;
      }
  }
}

---

Quote:

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Originally Posted by quzah

No, no braces are missing. They just need to indent so you can see it:

---

Ack, my bad. I didn't read those statements right at all. Looking back I'm not even sure what I read.

You need to drink more. That way you always have an excuse. Either you weren't drinking enough, or you've had too much. See? It's a win-win-win scenario!


Quzah.