REDUCE(result); \
} while(0)
- if (Py_SIZE(b) <= FIVEARY_CUTOFF) {
+ i = Py_SIZE(b);
+ digit bi = i ? b->ob_digit[i-1] : 0;
+ digit bit;
+ if (i <= 1 && bi <= 3) {
+ /* aim for minimal overhead */
+ if (bi >= 2) {
+ MULT(a, a, z);
+ if (bi == 3) {
+ MULT(z, a, z);
+ }
+ }
+ else if (bi == 1) {
+ /* Multiplying by 1 serves two purposes: if `a` is of an int
+ * subclass, makes the result an int (e.g., pow(False, 1) returns
+ * 0 instead of False), and potentially reduces `a` by the modulus.
+ */
+ MULT(a, z, z);
+ }
+ /* else bi is 0, and z==1 is correct */
+ }
+ else if (i <= FIVEARY_CUTOFF) {
/* Left-to-right binary exponentiation (HAC Algorithm 14.79) */
/* http://www.cacr.math.uwaterloo.ca/hac/about/chap14.pdf */
- for (i = Py_SIZE(b) - 1; i >= 0; --i) {
- digit bi = b->ob_digit[i];
- for (j = (digit)1 << (PyLong_SHIFT-1); j != 0; j >>= 1) {
+ /* Find the first significant exponent bit. Search right to left
+ * because we're primarily trying to cut overhead for small powers.
+ */
+ assert(bi); /* else there is no significant bit */
+ Py_INCREF(a);
+ Py_DECREF(z);
+ z = a;
+ for (bit = 2; ; bit <<= 1) {
+ if (bit > bi) { /* found the first bit */
+ assert((bi & bit) == 0);
+ bit >>= 1;
+ assert(bi & bit);
+ break;
+ }
+ }
+ for (--i, bit >>= 1;;) {
+ for (; bit != 0; bit >>= 1) {
MULT(z, z, z);
- if (bi & j)
+ if (bi & bit) {
MULT(z, a, z);
+ }
+ }
+ if (--i < 0) {
+ break;
}
+ bi = b->ob_digit[i];
+ bit = (digit)1 << (PyLong_SHIFT-1);
}
}
else {