mirror of https://github.com/python/cpython
Tweak the comments and formatting.
This commit is contained in:
Raymond Hettinger
parent
8f66a4a3db
commit
778d5cc4fb
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@ -311,61 +311,36 @@ FUNC1(tanh, tanh, 0,
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<http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/393090>,
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enhanced with the exact partials sum and roundoff from Mark
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Dickinson's post at <http://bugs.python.org/file10357/msum4.py>.
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See those links for more details, proofs and other references.
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See both of those for more details, proofs and other references.
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Note 1: IEEE 754R floating point semantics are assumed,
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but the current implementation does not re-establish special
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value semantics across iterations (i.e. handling -Inf + Inf).
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Note 1: IEEE 754 floating point format and semantics are assumed, but not
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explicitly maintained. The following rules may not apply:
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Note 2: No provision is made for intermediate overflow handling;
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therefore, sum([1e+308, 1e-308, 1e+308]) returns result 1e+308 while
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sum([1e+308, 1e+308, 1e-308]) raises an OverflowError due to the
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overflow of the first partial sum.
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1. if the summands include a NaN, return a NaN,
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Note 3: Aggressively optimizing compilers can potentially eliminate the
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residual values needed for accurate summation. For instance, the statements
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"hi = x + y; lo = y - (hi - x);" could be mis-transformed to
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"hi = x + y; lo = 0.0;" which defeats the computation of residuals.
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2. if the summands include infinities of both signs, raise ValueError,
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Note 4: A similar implementation is in Modules/cmathmodule.c.
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Be sure to update both when making changes.
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3. if the summands include infinities of only one sign, return infinity
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with that sign,
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4. otherwise (all summands are finite) if the result is infinite, raise
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OverflowError. The result can never be a NaN if all summands are
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finite.
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Note 2: the implementation below not include the intermediate overflow
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handling from Mark Dickinson's msum(). Therefore, sum([1e+308, 1e-308,
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1e+308]) returns result 1e+308, however sum([1e+308, 1e+308, 1e-308])
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raises an OverflowError due to intermediate overflow of the first
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partial sum.
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Note 3: aggressively optimizing compilers may eliminate the roundoff
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expressions critical for accurate summation. For example, the compiler
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may optimize the following expressions
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hi = x + y;
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lo = y - (hi - x);
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to
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hi = x + y;
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lo = 0.0;
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defeating the whole purpose. Using volatile variables and/or explicit
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assignment of critical subexpressions to a volatile variable should
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remedy the problem
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volatile double v; // Deter compiler from algebraically optimizing
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// this critical, intermediate value away
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hi = x + y;
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v = hi - x;
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lo = y - v;
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by forcing the compiler to compute the value for v. This may also help
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when subexpression are not computed with the full double precision.
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Note 4. the same summation functions may be in ./cmathmodule.c. Make
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sure to update both when making changes.
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Note 5: The signature of math.sum() differs from __builtin__.sum()
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because the start argument doesn't make sense in the context of
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accurate summation. Since the partials table is collapsed before
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returning a result, sum(seq2, start=sum(seq1)) may not equal the
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accurate result returned by sum(itertools.chain(seq1, seq2)).
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*/
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#define NUM_PARTIALS 32 /* initial partials array size, on stack */
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/* Extend the partials array p[] by doubling its size.
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*/
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static int /* non-zero on error */
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/* Extend the partials array p[] by doubling its size. */
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static int /* non-zero on error */
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_sum_realloc(double **p_ptr, Py_ssize_t n,
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double *ps, Py_ssize_t *m_ptr)
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{
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@ -383,7 +358,7 @@ _sum_realloc(double **p_ptr, Py_ssize_t n,
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else
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v = PyMem_Realloc(p, sizeof(double) * m);
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}
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if (v == NULL) { /* size overflow or no memory */
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if (v == NULL) { /* size overflow or no memory */
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PyErr_SetString(PyExc_MemoryError, "math sum partials");
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return 1;
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}
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@ -419,8 +394,9 @@ _sum_realloc(double **p_ptr, Py_ssize_t n,
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non-zero, non-special, non-overlapping and strictly increasing in
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magnitude, but possibly not all having the same sign.
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Depends on IEEE 754 arithmetic guarantees.
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*/
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Depends on IEEE 754 arithmetic guarantees and half-even rounding.
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*/
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static PyObject*
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math_sum(PyObject *self, PyObject *seq)
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{
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@ -434,8 +410,7 @@ math_sum(PyObject *self, PyObject *seq)
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PyFPE_START_PROTECT("sum", Py_DECREF(iter); return NULL)
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for(;;) { /* for x in iterable */
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/* some invariants */
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for(;;) { /* for x in iterable */
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assert(0 <= n && n <= m);
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assert((m == NUM_PARTIALS && p == ps) ||
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(m > NUM_PARTIALS && p != NULL));
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@ -444,28 +419,27 @@ math_sum(PyObject *self, PyObject *seq)
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if (item == NULL) {
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if (PyErr_Occurred())
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goto _sum_error;
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else
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break;
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break;
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}
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x = PyFloat_AsDouble(item);
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Py_DECREF(item);
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if (PyErr_Occurred())
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goto _sum_error;
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for (i = j = 0; j < n; j++) { /* for y in partials */
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for (i = j = 0; j < n; j++) { /* for y in partials */
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y = p[j];
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hi = x + y;
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lo = fabs(x) < fabs(y)
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? x - (hi - y) /* volatile */
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: y - (hi - x); /* volatile */
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? x - (hi - y)
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: y - (hi - x);
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if (lo != 0.0)
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p[i++] = lo;
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x = hi;
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}
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/* ps[i:] = [x] */
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n = i;
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n = i; /* ps[i:] = [x] */
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if (x != 0.0) {
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/* if non-finite, reset partials, effectively
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/* If non-finite, reset partials, effectively
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adding subsequent items without roundoff
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and yielding correct non-finite results,
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provided IEEE 754 rules are observed */
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@ -476,27 +450,27 @@ math_sum(PyObject *self, PyObject *seq)
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p[n++] = x;
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}
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}
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assert(n <= m);
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if (n > 0) {
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hi = p[--n];
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if (Py_IS_FINITE(hi)) {
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/* sum_exact(ps, hi) from the top, stop
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as soon as the sum becomes inexact */
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/* sum_exact(ps, hi) from the top, stop when the sum becomes inexact. */
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while (n > 0) {
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x = p[--n];
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y = hi;
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hi = x + y;
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assert(fabs(x) < fabs(y));
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lo = x - (hi - y); /* volatile */
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lo = x - (hi - y);
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if (lo != 0.0)
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break;
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}
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/* round correctly if necessary */
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/* Little dance to allow half-even rounding across multiple partials.
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Needed so that sum([1e-16, 1, 1e16]) will round-up to two instead
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of down to zero (the 1e16 makes the 1 slightly closer to two). */
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if (n > 0 && ((lo < 0.0 && p[n-1] < 0.0) ||
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(lo > 0.0 && p[n-1] > 0.0))) {
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y = lo * 2.0;
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x = hi + y; /* volatile */
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x = hi + y;
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if (y == (x - hi))
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hi = x;
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}
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@ -513,7 +487,6 @@ math_sum(PyObject *self, PyObject *seq)
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_sum_error:
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PyFPE_END_PROTECT(hi)
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Py_DECREF(iter);
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if (p != ps)
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PyMem_Free(p);
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@ -523,11 +496,9 @@ _sum_error:
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#undef NUM_PARTIALS
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PyDoc_STRVAR(math_sum_doc,
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"sum(sequence)\n\n\
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Return the full precision sum of a sequence of numbers.\n\
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When the sequence is empty, return zero.\n\n\
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For accurate results, IEEE 754 floating point format\n\
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and semantics and floating point radix 2 are required.");
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"sum(iterable)\n\n\
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Return an accurate floating point sum of values in the iterable.\n\
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Assumes IEEE-754 floating point arithmetic.");
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static PyObject *
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math_trunc(PyObject *self, PyObject *number)
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