diff Python/bltinmodule.c @ 10279:886ea255f381 legacy-trunk

Patch by Tim Peters to improve the range checks for range() and xrange(), especially for platforms where int and long are different sizes (so sys.maxint isn't actually the theoretical limit for the length of a list, but the largest C int is -- sys.maxint is the largest Python int, which is actually a C long).
author Guido van Rossum <guido@python.org>
date Tue, 23 Feb 1999 16:11:01 +0000
parents 94125a1343d8
children 1c6b10dea9f0
line wrap: on
line diff
--- a/Python/bltinmodule.c	Tue Feb 23 15:43:15 1999 +0000
+++ b/Python/bltinmodule.c	Tue Feb 23 16:11:01 1999 +0000
@@ -1395,13 +1395,47 @@
 equivalent to (x**y) % z, but may be more efficient (e.g. for longs).";
 
 
+/* Return number of items in range/xrange (lo, hi, step).  step > 0
+ * required.  Return a value < 0 if & only if the true value is too
+ * large to fit in a signed long.
+ */
+static long
+get_len_of_range(lo, hi, step)
+	long lo;
+	long hi;
+	long step;	/* must be > 0 */
+{
+	/* -------------------------------------------------------------
+	If lo >= hi, the range is empty.
+	Else if n values are in the range, the last one is
+	lo + (n-1)*step, which must be <= hi-1.  Rearranging,
+	n <= (hi - lo - 1)/step + 1, so taking the floor of the RHS gives
+	the proper value.  Since lo < hi in this case, hi-lo-1 >= 0, so
+	the RHS is non-negative and so truncation is the same as the
+	floor.  Letting M be the largest positive long, the worst case
+	for the RHS numerator is hi=M, lo=-M-1, and then
+	hi-lo-1 = M-(-M-1)-1 = 2*M.  Therefore unsigned long has enough
+	precision to compute the RHS exactly.
+	---------------------------------------------------------------*/
+	long n = 0;
+	if (lo < hi) {
+		unsigned long uhi = (unsigned long)hi;
+		unsigned long ulo = (unsigned long)lo;
+		unsigned long diff = uhi - ulo - 1;
+		n = (long)(diff / (unsigned long)step + 1);
+	}
+	return n;
+}
+
 static PyObject *
 builtin_range(self, args)
 	PyObject *self;
 	PyObject *args;
 {
 	long ilow = 0, ihigh = 0, istep = 1;
+	long bign;
 	int i, n;
+
 	PyObject *v;
 
 	if (PyTuple_Size(args) <= 1) {
@@ -1420,32 +1454,14 @@
 		PyErr_SetString(PyExc_ValueError, "zero step for range()");
 		return NULL;
 	}
-	/* A bit convoluted because this might overflow; due to Tim Peters */
-	if (istep > 0) {
-		if (ihigh <= ilow)
-			n = 0;
-		else {
-			unsigned long hi = (unsigned long)ihigh;
-			unsigned long lo = (unsigned long)ilow;
-			unsigned long diff = hi - lo - 1;
-			n = (long)(diff / istep + 1);
-		}
-	}
-	else {
-		/* But any errors in this branch are my own --Guido */
-		if (ihigh >= ilow)
-			n = 0;
-		else {
-			/* Swap lo and hi; use abs(istep) */
-			unsigned long hi = (unsigned long)ilow;
-			unsigned long lo = (unsigned long)ihigh;
-			unsigned long diff = hi - lo - 1;
-			n = (long)(diff / (-istep) + 1);
-		}
-	}
-	if (n < 0) {
+	if (istep > 0)
+		bign = get_len_of_range(ilow, ihigh, istep);
+	else
+		bign = get_len_of_range(ihigh, ilow, -istep);
+	n = (int)bign;
+	if (bign < 0 || (long)n != bign) {
 		PyErr_SetString(PyExc_OverflowError,
-				"range() has more than sys.maxint items");
+				"range() has too many items");
 		return NULL;
 	}
 	v = PyList_New(n);
@@ -1497,13 +1513,15 @@
 		PyErr_SetString(PyExc_ValueError, "zero step for xrange()");
 		return NULL;
 	}
-	/* XXX ought to check overflow of subtraction */
 	if (istep > 0)
-		n = (ihigh - ilow + istep - 1) / istep;
+		n = get_len_of_range(ilow, ihigh, istep);
 	else
-		n = (ihigh - ilow + istep + 1) / istep;
-	if (n < 0)
-		n = 0;
+		n = get_len_of_range(ihigh, ilow, -istep);
+	if (n < 0) {
+		PyErr_SetString(PyExc_OverflowError,
+				"xrange() has more than sys.maxint items");
+		return NULL;
+	}
 	return PyRange_New(ilow, n, istep, 1);
 }