dnl  Alpha mpn_modexact_1c_odd -- mpn exact remainder

dnl  Copyright 2003, 2004 Free Software Foundation, Inc.
dnl
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include(`../config.m4')


C      cycles/limb
C EV4:    47
C EV5:    30
C EV6:    15


C mp_limb_t mpn_modexact_1c_odd (mp_srcptr src, mp_size_t size, mp_limb_t d,
C                                mp_limb_t c)
C
C This code follows the "alternate" code in mpn/generic/mode1o.c,
C eliminating cbit+climb from the dependent chain.  This leaves,
C
C        ev4   ev5   ev6
C         1     3     1    subq   y = x - h
C        23    13     7    mulq   q = y * inverse
C        23    14     7    umulh  h = high (q * d)
C        --    --    --
C        47    30    15
C
C In each case, the load latency, loop control, and extra carry bit handling
C hide under the multiply latencies.  Those latencies are long enough that
C we don't need to worry about alignment or pairing to squeeze out
C performance.
C
C For the first limb, some of the loop code is broken out and scheduled back
C since it can be done earlier.
C
C   - The first ldq src[0] is near the start of the routine, for maximum
C     time from memory.
C
C   - The subq y=x-climb can be done without waiting for the inverse.
C
C   - The mulq y*inverse is replicated after the final subq for the inverse,
C     instead of branching to the mulq in the main loop.  On ev4 a branch
C     there would cost cycles, but we can hide them under the mulq latency.
C
C For the last limb, high<divisor is tested and if that's true a subtract
C and addback is done, as per the main mpn/generic/mode1o.c code.  This is a
C data-dependent branch, but we're waiting for umulh so any penalty should
C hide there.  The multiplies saved would be worth the cost anyway.
C
C Enhancements:
C
C For size==1, a plain division (done bitwise say) might be faster than
C calculating an inverse, the latter taking about 130 cycles on ev4 or 70 on
C ev5.  A call to gcc __remqu might be a possibility.

ASM_START()
PROLOGUE(mpn_modexact_1c_odd,gp)

	C r16	src
	C r17	size
	C r18	d
	C r19	c

	LEA(r0, modlimb_invert_table)
	srl	r18, 1, r20		C d >> 1

	and	r20, 127, r20		C idx = d>>1 & 0x7F

	addq	r0, r20, r21		C table + idx

	ldq_u	r20, 0(r21)		C table[idx] qword
	extbl	r20, r21, r20		C table[idx], inverse 8 bits

	mull	r20, r20, r7		C i*i
	addq	r20, r20, r20		C 2*i

	ldq	r2, 0(r16)		C x = s = src[0]
	lda	r17, -1(r17)		C size--
	clr	r0			C initial cbit=0

	mull	r7, r18, r7		C i*i*d

	subq	r20, r7, r20		C 2*i-i*i*d, inverse 16 bits

	mull	r20, r20, r7		C i*i
	addq	r20, r20, r20		C 2*i

	mull	r7, r18, r7		C i*i*d

	subq	r20, r7, r20		C 2*i-i*i*d, inverse 32 bits

	mulq	r20, r20, r7		C i*i
	addq	r20, r20, r20		C 2*i

	mulq	r7, r18, r7		C i*i*d
	subq	r2, r19, r3		C y = x - climb

	subq	r20, r7, r20		C inv = 2*i-i*i*d, inverse 64 bits

ASSERT(r7, C should have d*inv==1 mod 2^64
`	mulq	r18, r20, r7
	cmpeq	r7, 1, r7')

	mulq	r3, r20, r4		C first q = y * inv

	beq	r17, L(one)		C if size==1
	br	L(entry)


L(top):
	C r0	cbit
	C r16	src, incrementing
	C r17	size, decrementing
	C r18	d
	C r19	climb
	C r20	inv

	ldq	r1, 0(r16)		C s = src[i]
	subq	r1, r0, r2		C x = s - cbit
	cmpult	r1, r0, r0		C new cbit = s < cbit

	subq	r2, r19, r3		C y = x - climb

	mulq	r3, r20, r4		C q = y * inv
L(entry):
	cmpult	r2, r19, r5		C cbit2 = x < climb
	addq	r5, r0, r0		C cbit += cbit2
	lda	r16, 8(r16)		C src++
	lda	r17, -1(r17)		C size--

	umulh	r4, r18, r19		C climb = q * d
	bne	r17, L(top)		C while 2 or more limbs left



	C r0	cbit
	C r18	d
	C r19	climb
	C r20	inv

	ldq	r1, 0(r16)		C s = src[size-1] high limb

	cmpult	r1, r18, r2		C test high<divisor
	bne	r2, L(skip)		C skip if so

	C can't skip a division, repeat loop code

	subq	r1, r0, r2		C x = s - cbit
	cmpult	r1, r0, r0		C new cbit = s < cbit

	subq	r2, r19, r3		C y = x - climb

	mulq	r3, r20, r4		C q = y * inv
L(one):
	cmpult	r2, r19, r5		C cbit2 = x < climb
	addq	r5, r0, r0		C cbit += cbit2

	umulh	r4, r18, r19		C climb = q * d

	addq	r19, r0, r0		C return climb + cbit
	ret	r31, (r26), 1


	ALIGN(8)
L(skip):
	C with high<divisor, the final step can be just (cbit+climb)-s and
	C an addback of d if that underflows

	addq	r19, r0, r19		C c = climb + cbit

	subq	r19, r1, r2		C c - s
	cmpult	r19, r1, r3		C c < s

	addq	r2, r18, r0		C return c-s + divisor

	cmoveq	r3, r2, r0		C return c-s if no underflow
	ret	r31, (r26), 1

EPILOGUE()
ASM_END()
