Lines Matching defs:asize
3499 Py_ssize_t asize = Py_ABS(Py_SIZE(a));
3521 if (asize > bsize) {
3526 i = asize;
3527 asize = bsize;
3533 if (asize <= i) {
3534 if (asize == 0)
3546 if (2 * asize <= bsize)
3563 * 1. Allocate result space (asize + bsize digits: that's always
3579 ret = _PyLong_New(asize + bsize);
3672 3. asize <= bsize
3673 4. Since we call k_lopsided_mul if asize*2 <= bsize, asize*2 > bsize in this
3674 routine, so asize > bsize/2 >= f(bsize/2) in this routine.
3676 We allocated asize + bsize result digits, and add t3 into them at an offset
3677 of shift. This leaves asize+bsize-shift allocated digit positions for t3
3678 to fit into, = (by #1 and #2) asize + f(bsize/2) + c(bsize/2) - f(bsize/2) =
3679 asize + c(bsize/2) available digit positions.
3684 If asize == bsize, ah has c(bsize/2) digits, else ah has at most f(bsize/2)
3686 most (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 1 bit.
3690 c(bsize/2) + (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 2 bits
3692 and we have asize + c(bsize/2) available digit positions. We need to show
3694 the question reduces to whether asize digits is enough to hold
3695 (asize == bsize ? c(bsize/2) : f(bsize/2)) digits + 2 bits. If asize < bsize,
3696 then we're asking whether asize digits >= f(bsize/2) digits + 2 bits. By #4,
3697 asize is at least f(bsize/2)+1 digits, so this in turn reduces to whether 1
3699 asize == bsize, then we're asking whether bsize digits is enough to hold
3720 const Py_ssize_t asize = Py_ABS(Py_SIZE(a));
3726 assert(asize > KARATSUBA_CUTOFF);
3727 assert(2 * asize <= bsize);
3730 ret = _PyLong_New(asize + bsize);
3736 bslice = _PyLong_New(asize);
3743 const Py_ssize_t nbtouse = Py_MIN(bsize, asize);