📄 lb1spc.asm

📁 gcc库的原代码,对编程有很大帮助.
💻 ASM
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/* This is an assembly language implementation of libgcc1.c for the sparc   processor.   These routines are derived from the Sparc Architecture Manual, version 8,   slightly edited to match the desired calling convention, and also to   optimize them for our purposes.  */#ifdef L_mulsi3.text	.align 4	.global .umul	.proc 4.umul:	or	%o0, %o1, %o4	! logical or of multiplier and multiplicand	mov	%o0, %y		! multiplier to Y register	andncc	%o4, 0xfff, %o5	! mask out lower 12 bits	be	mul_shortway	! can do it the short way	andcc	%g0, %g0, %o4	! zero the partial product and clear NV cc	!	! long multiply	!	mulscc	%o4, %o1, %o4	! first iteration of 33	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	! 32nd iteration	mulscc	%o4, %g0, %o4	! last iteration only shifts	! the upper 32 bits of product are wrong, but we do not care	retl	rd	%y, %o0	!	! short multiply	!mul_shortway:	mulscc	%o4, %o1, %o4	! first iteration of 13	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	mulscc	%o4, %o1, %o4	! 12th iteration	mulscc	%o4, %g0, %o4	! last iteration only shifts	rd	%y, %o5	sll	%o4, 12, %o4	! left shift partial product by 12 bits	srl	%o5, 20, %o5	! right shift partial product by 20 bits	retl	or	%o5, %o4, %o0	! merge for true product#endif#ifdef L_divsi3.text	.align 4	.global	.udiv	.proc 4.udiv:	save	%sp, -64, %sp	b	divide	mov	0, %i2		! result always positive	.global .div	.proc 4.div:	save	%sp, -64, %sp	orcc	%i1, %i0, %g0	! is either operand negative	bge	divide		! if not, skip this junk	xor	%i1, %i0, %i2	! record sign of result in sign of %i2	tst	%i1	bge	2f	tst	%i0	! %i1 < 0	bge	divide	neg	%i12:	! %i0 < 0	neg	%i0	!	FALL THROUGHdivide:	! Compute size of quotient, scale comparand.	orcc	%i1, %g0, %l1		! movcc %i1, %l1	te	2			! if %i1 = 0	mov	%i0, %i3	mov	0, %i2	sethi	%hi(1<<(32-2-1)), %l3	cmp 	%i3, %l3	blu	not_really_big	mov	0, %l0	!	! Here, the %i0 is >= 2^(31-3) or so.  We must be careful here,	! as our usual 3-at-a-shot divide step will cause overflow and havoc.	! The total number of bits in the result here is 3*%l0+%l4, where	! %l4 <= 3.	! Compute %l0 in an unorthodox manner: know we need to Shift %l1 into	! the top decade: so do not even bother to compare to %i3.1:	cmp	%l1, %l3	bgeu	3f	mov	1, %l4	sll	%l1, 3, %l1	b	1b	inc	%l0	!	! Now compute %l4	!2:	addcc	%l1, %l1, %l1	bcc	not_too_big	add	%l4, 1, %l4	!	! We are here if the %i1 overflowed when Shifting.	! This means that %i3 has the high-order bit set.	! Restore %l1 and subtract from %i3.	sll	%l3, 2, %l3	srl	%l1, 1, %l1	add	%l1, %l3, %l1	b	do_single_div	dec	%l4not_too_big:3:	cmp	%l1, %i3	blu	2b	nop	be	do_single_div	nop	! %l1 > %i3: went too far: back up 1 step	! 	srl	%l1, 1, %l1	!	dec	%l4	! do single-bit divide steps	!	! We have to be careful here.  We know that %i3 >= %l1, so we can do the	! first divide step without thinking.  BUT, the others are conditional,	! and are only done if %i3 >= 0.  Because both %i3 and %l1 may have the	! high-order bit set in the first step, just falling into the regular	! division loop will mess up the first time around.	! So we unroll slightly...do_single_div:	deccc	%l4	bl	end_regular_divide	nop	sub	%i3, %l1, %i3	mov	1, %i2	b	end_single_divloop	nopsingle_divloop:	sll	%i2, 1, %i2	bl	1f	srl	%l1, 1, %l1	! %i3 >= 0	sub	%i3, %l1, %i3	b	2f	inc	%i21:	! %i3 < 0	add	%i3, %l1, %i3	dec	%i2end_single_divloop:2:	deccc	%l4	bge	single_divloop	tst	%i3	b	end_regular_divide	nopnot_really_big:1:	sll	%l1, 3, %l1	cmp	%l1, %i3	bleu	1b	inccc	%l0	be	got_result	dec	%l0do_regular_divide:	! Do the main division iteration	tst	%i3	! Fall through into divide loopdivloop:	sll	%i2, 3, %i2	! depth 1, accumulated bits 0	bl	L.1.8	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 2, accumulated bits 1	bl	L.2.9	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 3, accumulated bits 3	bl	L.3.11	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (3*2+1), %i2L.3.11:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (3*2-1), %i2L.2.9:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 3, accumulated bits 1	bl	L.3.9	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (1*2+1), %i2L.3.9:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (1*2-1), %i2L.1.8:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 2, accumulated bits -1	bl	L.2.7	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 3, accumulated bits -1	bl	L.3.7	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (-1*2+1), %i2L.3.7:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (-1*2-1), %i2L.2.7:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 3, accumulated bits -3	bl	L.3.5	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (-3*2+1), %i2L.3.5:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (-3*2-1), %i2end_regular_divide:9:	deccc	%l0	bge	divloop	tst	%i3	bge	got_result	nop	! non-restoring fixup here	dec	%i2got_result:	tst	%i2	bge	1f	restore	! answer < 0	retl		! leaf-routine return	neg	%o2, %o0	! quotient <- -%i21:	retl		! leaf-routine return	mov	%o2, %o0	! quotient <- %i2#endif#ifdef L_modsi3.text	.align 4	.global	.urem	.proc 4.urem:	save	%sp, -64, %sp	b	divide	mov	0, %i2		! result always positive	.global .rem	.proc 4.rem:	save	%sp, -64, %sp	orcc	%i1, %i0, %g0	! is either operand negative	bge	divide		! if not, skip this junk	mov	%i0, %i2	! record sign of result in sign of %i2	tst	%i1	bge	2f	tst	%i0	! %i1 < 0	bge	divide	neg	%i12:	! %i0 < 0	neg	%i0	!	FALL THROUGHdivide:	! Compute size of quotient, scale comparand.	orcc	%i1, %g0, %l1		! movcc %i1, %l1	te	2			! if %i1 = 0	mov	%i0, %i3	mov	0, %i2	sethi	%hi(1<<(32-2-1)), %l3	cmp 	%i3, %l3	blu	not_really_big	mov	0, %l0	!	! Here, the %i0 is >= 2^(31-3) or so.  We must be careful here,	! as our usual 3-at-a-shot divide step will cause overflow and havoc.	! The total number of bits in the result here is 3*%l0+%l4, where	! %l4 <= 3.	! Compute %l0 in an unorthodox manner: know we need to Shift %l1 into	! the top decade: so do not even bother to compare to %i3.1:	cmp	%l1, %l3	bgeu	3f	mov	1, %l4	sll	%l1, 3, %l1	b	1b	inc	%l0	!	! Now compute %l4	!2:	addcc	%l1, %l1, %l1	bcc	not_too_big	add	%l4, 1, %l4	!	! We are here if the %i1 overflowed when Shifting.	! This means that %i3 has the high-order bit set.	! Restore %l1 and subtract from %i3.	sll	%l3, 2, %l3	srl	%l1, 1, %l1	add	%l1, %l3, %l1	b	do_single_div	dec	%l4not_too_big:3:	cmp	%l1, %i3	blu	2b	nop	be	do_single_div	nop	! %l1 > %i3: went too far: back up 1 step	! 	srl	%l1, 1, %l1	!	dec	%l4	! do single-bit divide steps	!	! We have to be careful here.  We know that %i3 >= %l1, so we can do the	! first divide step without thinking.  BUT, the others are conditional,	! and are only done if %i3 >= 0.  Because both %i3 and %l1 may have the	! high-order bit set in the first step, just falling into the regular	! division loop will mess up the first time around.	! So we unroll slightly...do_single_div:	deccc	%l4	bl	end_regular_divide	nop	sub	%i3, %l1, %i3	mov	1, %i2	b	end_single_divloop	nopsingle_divloop:	sll	%i2, 1, %i2	bl	1f	srl	%l1, 1, %l1	! %i3 >= 0	sub	%i3, %l1, %i3	b	2f	inc	%i21:	! %i3 < 0	add	%i3, %l1, %i3	dec	%i2end_single_divloop:2:	deccc	%l4	bge	single_divloop	tst	%i3	b	end_regular_divide	nopnot_really_big:1:	sll	%l1, 3, %l1	cmp	%l1, %i3	bleu	1b	inccc	%l0	be	got_result	dec	%l0do_regular_divide:	! Do the main division iteration	tst	%i3	! Fall through into divide loopdivloop:	sll	%i2, 3, %i2	! depth 1, accumulated bits 0	bl	L.1.8	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 2, accumulated bits 1	bl	L.2.9	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 3, accumulated bits 3	bl	L.3.11	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (3*2+1), %i2L.3.11:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (3*2-1), %i2L.2.9:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 3, accumulated bits 1	bl	L.3.9	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (1*2+1), %i2L.3.9:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (1*2-1), %i2L.1.8:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 2, accumulated bits -1	bl	L.2.7	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	! depth 3, accumulated bits -1	bl	L.3.7	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (-1*2+1), %i2L.3.7:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (-1*2-1), %i2L.2.7:	! remainder is negative	addcc	%i3,%l1,%i3	! depth 3, accumulated bits -3	bl	L.3.5	srl	%l1,1,%l1	! remainder is positive	subcc	%i3,%l1,%i3	b	9f	add	%i2, (-3*2+1), %i2L.3.5:	! remainder is negative	addcc	%i3,%l1,%i3	b	9f	add	%i2, (-3*2-1), %i2end_regular_divide:9:	deccc	%l0	bge	divloop	tst	%i3	bge	got_result	nop	! non-restoring fixup here	add	%i3, %i1, %i3got_result:	tst	%i2	bge	1f	restore	! answer < 0	retl		! leaf-routine return	neg	%o3, %o0	! remainder <- -%i31:	retl		! leaf-routine return	mov	%o3, %o0	! remainder <- %i3#endif
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