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Commit a3c01ce6 authored by Tim Peters's avatar Tim Peters
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SF bug #488480: integer multiply to return -max_int-1. 

int_mul():  new and vastly simpler overflow checking.  Whether it's
faster or slower will likely vary across platforms, favoring boxes
with fast floating point.  OTOH, we no longer have to worry about
people shipping broken LONG_BIT definitions <0.9 wink>.
parent d2a557e5
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Showing with 89 additions and 127 deletions
Lib/test/test_types.py
View file @ a3c01ce6
......@@ -71,6 +71,32 @@ if not -24 < -12: raise TestFailed, 'int op'
xsize, ysize, zsize = 238, 356, 4
if not (xsize*ysize*zsize == zsize*xsize*ysize == 338912):
raise TestFailed, 'int mul commutativity'
# And another.
m = -sys.maxint - 1
for divisor in 1, 2, 4, 8, 16, 32:
j = m / divisor
prod = divisor * j
if prod != m:
raise TestFailed, "%r * %r == %r != %r" % (divisor, j, prod, m)
if type(prod) is not int:
raise TestFailed, ("expected type(prod) to be int, not %r" %
type(prod))
# Check for expected * overflow to long.
for divisor in 1, 2, 4, 8, 16, 32:
j = m / divisor - 1
prod = divisor * j
if type(prod) is not long:
raise TestFailed, ("expected type(%r) to be long, not %r" %
(prod, type(prod)))
# Check for expected * overflow to long.
m = sys.maxint
for divisor in 1, 2, 4, 8, 16, 32:
j = m / divisor + 1
prod = divisor * j
if type(prod) is not long:
raise TestFailed, ("expected type(%r) to be long, not %r" %
(prod, type(prod)))
print '6.4.2 Long integers'
if 12L + 24L != 36L: raise TestFailed, 'long op'
if 12L + (-24L) != -12L: raise TestFailed, 'long op'
......
Objects/intobject.c
View file @ a3c01ce6
......@@ -148,16 +148,16 @@ PyInt_AsLong(register PyObject *op)
PyNumberMethods *nb;
PyIntObject *io;
long val;
if (op && PyInt_Check(op))
return PyInt_AS_LONG((PyIntObject*) op);
if (op == NULL || (nb = op->ob_type->tp_as_number) == NULL ||
nb->nb_int == NULL) {
PyErr_SetString(PyExc_TypeError, "an integer is required");
return -1;
}
io = (PyIntObject*) (*nb->nb_int) (op);
if (io == NULL)
return -1;
......@@ -166,10 +166,10 @@ PyInt_AsLong(register PyObject *op)
"nb_int should return int object");
return -1;
}
val = PyInt_AS_LONG(io);
Py_DECREF(io);
return val;
}
......@@ -219,7 +219,7 @@ PyObject *
PyInt_FromUnicode(Py_UNICODE *s, int length, int base)
{
char buffer[256];
if (length >= sizeof(buffer)) {
PyErr_SetString(PyExc_ValueError,
"int() literal too large to convert");
......@@ -312,39 +312,38 @@ int_sub(PyIntObject *v, PyIntObject *w)
}
/*
Integer overflow checking used to be done using a double, but on 64
bit machines (where both long and double are 64 bit) this fails
because the double doesn't have enough precision. John Tromp suggests
the following algorithm:
Suppose again we normalize a and b to be nonnegative.
Let ah and al (bh and bl) be the high and low 32 bits of a (b, resp.).
Now we test ah and bh against zero and get essentially 3 possible outcomes.
1) both ah and bh > 0 : then report overflow
2) both ah and bh = 0 : then compute a*b and report overflow if it comes out
negative
3) ah > 0 and bh = 0 : compute ah*bl and report overflow if it's >= 2^31
compute al*bl and report overflow if it's negative
add (ah*bl)<<32 to al*bl and report overflow if
it's negative
In case of no overflow the result is then negated if necessary.
The majority of cases will be 2), in which case this method is the same as
what I suggested before. If multiplication is expensive enough, then the
other method is faster on case 3), but also more work to program, so I
guess the above is the preferred solution.
Integer overflow checking for * is painful: Python tried a couple ways, but
they didn't work on all platforms, or failed in endcases (a product of
-sys.maxint-1 has been a particular pain).
Here's another way:
The native long product x*y is either exactly right or *way* off, being
just the last n bits of the true product, where n is the number of bits
in a long (the delivered product is the true product plus i*2**n for
some integer i).
The native double product (double)x * (double)y is subject to three
rounding errors: on a sizeof(long)==8 box, each cast to double can lose
info, and even on a sizeof(long)==4 box, the multiplication can lose info.
But, unlike the native long product, it's not in *range* trouble: even
if sizeof(long)==32 (256-bit longs), the product easily fits in the
dynamic range of a double. So the leading 50 (or so) bits of the double
product are correct.
We check these two ways against each other, and declare victory if they're
approximately the same. Else, because the native long product is the only
one that can lose catastrophic amounts of information, it's the native long
product that must have overflowed.
*/
static PyObject *
int_mul(PyObject *v, PyObject *w)
{
long a, b, ah, bh, x, y;
int s = 1;
long a, b;
long longprod; /* a*b in native long arithmetic */
double doubled_longprod; /* (double)longprod */
double doubleprod; /* (double)a * (double)b */
if (!PyInt_Check(v) &&
v->ob_type->tp_as_sequence &&
......@@ -363,97 +362,34 @@ int_mul(PyObject *v, PyObject *w)
CONVERT_TO_LONG(v, a);
CONVERT_TO_LONG(w, b);
ah = a >> (LONG_BIT/2);
bh = b >> (LONG_BIT/2);
/* Quick test for common case: two small positive ints */
if (ah == 0 && bh == 0) {
x = a*b;
if (x < 0)
goto bad;
return PyInt_FromLong(x);
}
/* Arrange that a >= b >= 0 */
if (a < 0) {
a = -a;
if (a < 0) {
/* Largest negative */
if (b == 0 || b == 1) {
x = a*b;
goto ok;
}
else
goto bad;
}
s = -s;
ah = a >> (LONG_BIT/2);
}
if (b < 0) {
b = -b;
if (b < 0) {
/* Largest negative */
if (a == 0 || (a == 1 && s == 1)) {
x = a*b;
goto ok;
}
else
goto bad;
}
s = -s;
bh = b >> (LONG_BIT/2);
}
/* 1) both ah and bh > 0 : then report overflow */
if (ah != 0 && bh != 0)
goto bad;
/* 2) both ah and bh = 0 : then compute a*b and report
overflow if it comes out negative */
if (ah == 0 && bh == 0) {
x = a*b;
if (x < 0)
goto bad;
return PyInt_FromLong(x*s);
}
if (a < b) {
/* Swap */
x = a;
a = b;
b = x;
ah = bh;
/* bh not used beyond this point */
longprod = a * b;
doubleprod = (double)a * (double)b;
doubled_longprod = (double)longprod;
/* Fast path for normal case: small multiplicands, and no info
is lost in either method. */
if (doubled_longprod == doubleprod)
return PyInt_FromLong(longprod);
/* Somebody somewhere lost info. Close enough, or way off? Note
that a != 0 and b != 0 (else doubled_longprod == doubleprod == 0).
The difference either is or isn't significant compared to the
true value (of which doubleprod is a good approximation).
*/
{
const double diff = doubled_longprod - doubleprod;
const double absdiff = diff >= 0.0 ? diff : -diff;
const double absprod = doubleprod >= 0.0 ? doubleprod :
-doubleprod;
/* absdiff/absprod <= 1/32 iff
32 * absdiff <= absprod -- 5 good bits is "close enough" */
if (32.0 * absdiff <= absprod)
return PyInt_FromLong(longprod);
else if (err_ovf("integer multiplication"))
return NULL;
else
return PyLong_Type.tp_as_number->nb_multiply(v, w);
}
/* 3) ah > 0 and bh = 0 : compute ah*bl and report overflow if
it's >= 2^31
compute al*bl and report overflow if it's negative
add (ah*bl)<<32 to al*bl and report overflow if
it's negative
(NB b == bl in this case, and we make a = al) */
y = ah*b;
if (y >= (1L << (LONG_BIT/2 - 1)))
goto bad;
a &= (1L << (LONG_BIT/2)) - 1;
x = a*b;
if (x < 0)
goto bad;
x += y << (LONG_BIT/2);
if (x < 0)
goto bad;
ok:
return PyInt_FromLong(x * s);
bad:
if (err_ovf("integer multiplication"))
return NULL;
return PyLong_Type.tp_as_number->nb_multiply(v, w);
}
/* Return type of i_divmod */
......@@ -468,7 +404,7 @@ i_divmod(register long x, register long y,
long *p_xdivy, long *p_xmody)
{
long xdivy, xmody;
if (y == 0) {
PyErr_SetString(PyExc_ZeroDivisionError,
"integer division or modulo by zero");
......@@ -623,7 +559,7 @@ int_pow(PyIntObject *v, PyIntObject *w, PyIntObject *z)
ix = 1;
while (iw > 0) {
prev = ix; /* Save value for overflow check */
if (iw & 1) {
if (iw & 1) {
ix = ix*temp;
if (temp == 0)
break; /* Avoid ix / 0 */
......@@ -666,7 +602,7 @@ int_pow(PyIntObject *v, PyIntObject *w, PyIntObject *z)
}
}
return PyInt_FromLong(ix);
}
}
static PyObject *
int_neg(PyIntObject *v)
......
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