Lines Matching defs:bit
94 /* Suppose the exponent has bit length e. All ways of doing this
96 * each bit set. In a k-ary method with window width w, a multiply
110 * access to the bit length, though, so call it 60, which is a
111 * multiple of a long digit's max bit length (15 or 30 so far).
434 * then. The bit pattern for the largest positive signed long is
857 /* 2's-comp is a bit tricky here, e.g. 0xff00 == -0x0100, so
882 8-bit bytes and (probably) 15-bit Python digits.*/
991 * make sure at least one sign bit gets stored. */
1031 just above didn't get to ensure there's a sign bit, and the
1032 loop below wouldn't add one either. Make sure a sign bit
1043 /* Fill remaining bytes with copies of the sign bit. */
1160 size_t t; /* unsigned so >> doesn't propagate sign bit */
1569 borrow &= 1; /* keep only 1 sign bit */
1626 generated by the compiler despite us working with 30-bit digit values.
2138 /* Table of digit values for 8-bit string -> integer conversion.
2398 is a little bit larger than an exact integer, but due to roundoff errors (in
3022 bit of x to be 'sticky' for this shift: if any of the bits
3023 shifted out was nonzero, we set the least significant bit
3264 borrow &= 1; /* Keep only one sign bit */
3270 borrow &= 1; /* Keep only one sign bit */
3458 viewing the shift as being by digits. The sign bit is ignored, and
3682 at most c(bsize/2) digits + 1 bit.
3686 most (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 1 bit.
4010 at the end. Now write a_bits and b_bits for the bit lengths of a
4068 significant bit of x (above those to be rounded) is 1.
4097 the x87 FPU set to 64-bit precision. */
4523 digit bit;
4545 /* Find the first significant exponent bit. Search right to left
4548 assert(bi); /* else there is no significant bit */
4552 for (bit = 2; ; bit <<= 1) {
4553 if (bit > bi) { /* found the first bit */
4554 assert((bi & bit) == 0);
4555 bit >>= 1;
4556 assert(bi & bit);
4560 for (--i, bit >>= 1;;) {
4561 for (; bit != 0; bit >>= 1) {
4563 if (bi & bit) {
4571 bit = (digit)1 << (PyLong_SHIFT-1);
4591 * into `pending`, starting with the next 1 bit. The current bit
4619 const int bit = (bi >> j) & 1;
4620 pending = (pending << 1) | bit;
5773 /* Each digit has up to PyLong_SHIFT ones, so the accumulated bit count