|
idth);
}
if (__COMPARE(context, mid, hi) > 0) {
swap(mid, hi, width);
}
/* We now wish to partition the array into three pieces, one consisting
of elements <= partition element, one of elements equal to the
partition element, and one of elements > than it. This is done
below; comments indicate conditions established at every step. */
loguy = lo;
higuy = hi;
/* Note that higuy decreases and loguy increases on every iteration,
so loop must terminate. */
for (;;) {
/* lo <= loguy < hi, lo < higuy <= hi,
A[i] <= A[mid] for lo <= i <= loguy,
A[i] > A[mid] for higuy <= i < hi,
A[hi] >= A[mid] */
/* The doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same
value for both pointers. */
if (mid > loguy) {
do {
loguy += width;
} while (loguy < mid && __COMPARE(context, loguy, mid) <= 0);
}
if (mid <= loguy) {
do {
loguy += width;
} while (loguy <= hi && __COMPARE(context, loguy, mid) <= 0);
}
/* lo < loguy <= hi+1, A[i] <= A[mid] for lo <= i < loguy,
either loguy > hi or A[loguy] > A[mid] */
do {
higuy -= width;
} while (higuy > mid && __COMPARE(context, higuy, mid) > 0);
/* lo <= higuy < hi, A[i] > A[mid] for higuy < i < hi,
either higuy == lo or A[higuy] <= A[mid] */
if (higuy < loguy)
break;
/* if loguy > hi or higuy == lo, then we would have exited, so
A[loguy] > A[mid], A[higuy] <= A[mid],
loguy <= hi, higuy > lo */
swap(loguy, higuy, width);
/* If the partition element was moved, follow it. Only need
to check for mid == higuy, since before the swap,
A[loguy] > A[mid] implies loguy != mid. */
if (mid == higuy)
mid = loguy;
/* A[loguy] <= A[mid], A[higuy] > A[mid]; so condition at top
of loop is re-established */
}
/* A[i] <= A[mid] for lo <= i < loguy,
A[i] > A[mid] for higuy < i < hi,
A[hi] >= A[mid]
higuy < loguy
implying:
higuy == loguy-1
or higuy == hi - 1, loguy == hi + 1, A[hi] == A[mid] */
/* Find adjacent elements equal to the partition element. The
doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same value
for both pointers. */
higuy += width;
if (mid < higuy) {
do {
higuy -= width;
} while (higuy > mid && __COMPARE(context, higuy, mid) == 0);
}
if (mid >= higuy) {
do {
higuy -= width;
} while (higuy > lo && __COMPARE(context, higuy, mid) == 0);
}
/* OK, now we have the following:
higuy < loguy
lo <= higuy <= hi
A[i] <= A[mid] for lo <= i <= higuy
A[i] == A[mid] for higuy < i < loguy
A[i] > A[mid] for loguy <= i < hi
A[hi] >= A[mid] */
/* We've finished the partition, now we want to sort the subarrays
[lo, higuy] and [loguy, hi].
We do the smaller one first to minimize stack usage.
We only sort arrays of length 2 or more.*/
if ( higuy - lo >= hi - loguy ) {
if (lo < higuy) {
lostk[stkptr] = lo;
histk[stkptr] = higuy;
++stkptr;
} /* save big recursion for later */
if (loguy < hi) {
lo = loguy;
goto recurse; /* do small recursion */
}
}
else {
if (loguy < hi) {
lostk[stkptr] = loguy;
histk[stkptr] = hi;
++stkptr; /* save big recursion for later */
}
if (lo < higuy) {
hi = higuy;
goto recurse; /* do small recursion */
}
}
}
/* We have sorted the array, except for any pending sorts on the stack.
Check if there are any, and do them. */
--stkptr;
if (stkptr >= 0) {
lo = lostk[stkptr]; |