Issue #9599: Further accuracy tweaks to loghelper. For an integer n that's...

Issue #9599:  Further accuracy tweaks to loghelper.  For an integer n that's small enough to be converted to a float without overflow, log(n) is now computed as log(float(n)), and similarly for log10.

Lib/test/test_math.py

@@ -641,8 +641,12 @@ class MathTests(unittest.TestCase):
         self.ftest('log(32,2)', math.log(32,2), 5)
         self.ftest('log(10**40, 10)', math.log(10**40, 10), 40)
         self.ftest('log(10**40, 10**20)', math.log(10**40, 10**20), 2)
-        self.assertEquals(math.log(INF), INF)
+        self.ftest('log(10**1000)', math.log(10**1000),
+                   2302.5850929940457)
+        self.assertRaises(ValueError, math.log, -1.5)
+        self.assertRaises(ValueError, math.log, -10**1000)
         self.assertRaises(ValueError, math.log, NINF)
+        self.assertEquals(math.log(INF), INF)
         self.assertTrue(math.isnan(math.log(NAN)))
 
     def testLog1p(self):
@@ -655,8 +659,11 @@ class MathTests(unittest.TestCase):
         self.ftest('log10(0.1)', math.log10(0.1), -1)
         self.ftest('log10(1)', math.log10(1), 0)
         self.ftest('log10(10)', math.log10(10), 1)
-        self.assertEquals(math.log(INF), INF)
+        self.ftest('log10(10**1000)', math.log10(10**1000), 1000.0)
+        self.assertRaises(ValueError, math.log10, -1.5)
+        self.assertRaises(ValueError, math.log10, -10**1000)
         self.assertRaises(ValueError, math.log10, NINF)
+        self.assertEquals(math.log(INF), INF)
         self.assertTrue(math.isnan(math.log10(NAN)))
 
     def testModf(self):

Modules/mathmodule.c

@@ -1562,25 +1562,33 @@ loghelper(PyObject* arg, double (*func)(double), char *funcname)
 {
     /* If it is long, do it ourselves. */
     if (PyLong_Check(arg)) {
-        double x;
+        double x, result;
         Py_ssize_t e;
-        x = _PyLong_Frexp((PyLongObject *)arg, &e);
-        if (x == -1.0 && PyErr_Occurred())
-            return NULL;
-        if (x <= 0.0) {
+
+        /* Negative or zero inputs give a ValueError. */
+        if (Py_SIZE(arg) <= 0) {
             PyErr_SetString(PyExc_ValueError,
                             "math domain error");
             return NULL;
         }
-        /* Value is ~= x * 2**e, so the log ~= log(x) + log(2) * e.
-
-           It's slightly better to compute the log as log(2 * x) + log(2) * (e
-           - 1): then when 'arg' is a power of 2, 2**k say, this gives us 0.0 +
-           log(2) * k instead of log(0.5) + log(2)*(k+1), and so marginally
-           increases the chances of log(arg, 2) returning the correct result.
-        */
-        x = func(2.0 * x) + func(2.0) * (e - 1);
-        return PyFloat_FromDouble(x);
+
+        x = PyLong_AsDouble(arg);
+        if (x == -1.0 && PyErr_Occurred()) {
+            if (!PyErr_ExceptionMatches(PyExc_OverflowError))
+                return NULL;
+            /* Here the conversion to double overflowed, but it's possible
+               to compute the log anyway.  Clear the exception and continue. */
+            PyErr_Clear();
+            x = _PyLong_Frexp((PyLongObject *)arg, &e);
+            if (x == -1.0 && PyErr_Occurred())
+                return NULL;
+            /* Value is ~= x * 2**e, so the log ~= log(x) + log(2) * e. */
+            result = func(x) + func(2.0) * e;
+        }
+        else
+            /* Successfully converted x to a double. */
+            result = func(x);
+        return PyFloat_FromDouble(result);
     }
 
     /* Else let libm handle it by itself. */