Question

I have a problem understanding quicksort algorithm (the simplified version without pointers) from K&R. There is already a thorough explanation provided by Dave Gamble here explanation.
However I noticed that by starting with a slightly changed string we can obtain no swaps during many loops of the for loop.
Firstly the code:

void qsort(int v[], int left, int right)
{
    int i, last;

    void swap(int v[], int i, int j);

    if (left >= right) /* do nothing if array contains */
            return;         /* fewer than two elements */
    swap(v, left, (left + right)/2); /* move partition elem */
    last = left;                    /* to v[0] */
    for (i = left + 1; i <= right; i++) /* partition */
            if (v[i] < v[left])
                    swap(v, ++last, i);
    swap(v, left, last);                  /* restore partition elem */
    qsort(v, left, last-1);
    qsort(v, last+1, right);
}

Walkthrough in my opinion:

we start with CADBE; left=0; right=4; D is the pivot
so according to algorithm we swap D with C obtaining DACBE

last = left =0

i = 1 if ( v1 < v[0] ) it is true so we swap v1 (because last is incremented before operation) with v1 so nothing changes, last = 1, still having DACBE;

now i = 2 if ( v[2] < v[0] ) -> true so we swap v[2] with v[2] nothing changed again; last = 2

now i = 3 if ( v[3] < v[0] ) -> true so we swap v[3] with v[3] nothing changed AGAIN (!), last = 3

So apparently something is wrong, algorithm does nothing.
Your opinions appreciated very much. I must be wrong, authors are better than me ;D
Thanks in advance!

Answer 1

The loop goes from left + 1 up to and including right. When i=4, the test fails and last does not get incremented.

Then the recursive calls sort BACDE with left=0,right=2 and left=4,right=4. (Which is correct when D is the pivot.)