Thread: Pointers, Pointer Arithmetic and Arrays

I'm having a bit of trouble understanding pointers and arrays. As well as pointer arithmetic. The problem is as follows:

What will be the contents of the a array after following statements are executed?

Code:

#define N 10

int a[N] = {1 , 2, 3, 4, 5, 6, 7, 8, 9, 10};
int *p = &a[0], *q = &a[N - 1], temp;

while(p < q){
    temp = *p;
    *p++ = *q;
    *q-- = temp;
}

I know the above code will revers the array because temp is assigned to what p is pointing to, then p is incremented to advance in the elements of the array and assigned to what q is pointing to. Then q is decremented and assigned to temp (which hold the value of what p was initially pointing to).

I had similar problem on a quiz and I got the answer incorrect. The code is as follows:

Code:

#define N 10

int a[N] = {1 , 2, 3, 4, 5, 6, 7, 8, 9, 10};
int *p = &a[0], *q = &a[N - 1];

while(p < q){
    *p = *q;
    *q = *p;
    p++;
    q--;
}

I know p and q still get swapped in this situation. I think what tripped me up is the way p is incremented and q is decremented. Does the above code still reverse the array?

> What will be the contents of the a array after following statements are executed?
...
> I know the above code will revers the array because
Your analysis is correct.
So you know that the array will be 10,9,8,7,6,5,4,3,2,1 right?

> temp is assigned to what p is pointing to, then p is incremented
Yes.

> then p is incremented to advance in the elements of the array and assigned to what q is pointing to.
> Then q is decremented and assigned to temp (which hold the value of what p was initially pointing to).
No.
The assignments happen first, then the increment/decrement happens.
Check the position of the ++,-- in relation to the variable.

> I know p and q still get swapped in this situation
...
> I had similar problem on a quiz and I go the answer incorrect
So you don't know then.

Your 2nd example, there is no swapping.
It does however mirror the values in both halves of the array.

Then q is decremented and assigned to temp

temp is assigned to the address that q is pointing to, and then q is decremented.
Remember that with "post" increment or decrement, the increment/decrement expression has the old value even though the variable will ultimately be incremented.
So this:

Code:

    while (p < q){
        temp = *p;
        *p++ = *q;
        *q-- = temp;
    }

Is the same as this:

Code:

    while (p < q){
        temp = *p;
        *p = *q;
        ++p;
        *q = temp;
        --q;
    }

In your second version, since you don't use a temp variable to hold the initial value of p, after q is assigned to p, p equals q and therefore assigning it back to q just assigns q's original value back to q.

Code:

#include <stdio.h>
 
#define N 10
 
int main() {
 
    int a[N] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int *p = &a[0], *q = &a[N - 1], temp;
 
    while (p < q){
        temp = *p;
        *p++ = *q;
        *q-- = temp;
    }
 
    for (int i = 0; i < N; ++i)
        printf("%d ", a[i]);
    putchar('\n');
 
 
    int a2[N] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int *p2 = &a2[0], *q2 = &a2[N - 1];
 
    while (p2 < q2) {
        *p2 = *q2;
        *q2 = *p2;
        p2++;
        q2--;
    }
 
    for (int i = 0; i < N; ++i)
        printf("%d ", a2[i]);
    putchar('\n');
 
    return 0;
}
 
/*
Output:
10 9 8 7 6 5 4 3 2 1 
10 9 8 7 6 6 7 8 9 10 
*/

It takes seconds to just try it.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(void)
{
    #define N 10

int a[N] = {1 , 2, 3, 4, 5, 6, 7, 8, 9, 10};
int *p = &a[0], *q = &a[N - 1];

while(p < q){
    *p = *q;
    *q = *p;
    p++;
    q--;
}
    for ( int i = 0 ; i < N ; i++ ) {
        printf("%d %d\n", i, a[i] );
    }
   return (0);
}

$ gcc main.c
$ ./a.out 
0 10
1 9
2 8
3 7
4 6
5 6
6 7
7 8
8 9
9 10