diff options
Diffstat (limited to 'libc/sysdeps/linux/sparc/sparcv7/umul.S')
| -rw-r--r-- | libc/sysdeps/linux/sparc/sparcv7/umul.S | 153 | 
1 files changed, 153 insertions, 0 deletions
diff --git a/libc/sysdeps/linux/sparc/sparcv7/umul.S b/libc/sysdeps/linux/sparc/sparcv7/umul.S new file mode 100644 index 000000000..50b3157db --- /dev/null +++ b/libc/sysdeps/linux/sparc/sparcv7/umul.S @@ -0,0 +1,153 @@ +/* + * Unsigned multiply.  Returns %o0 * %o1 in %o1%o0 (i.e., %o1 holds the + * upper 32 bits of the 64-bit product). + * + * This code optimizes short (less than 13-bit) multiplies.  Short + * multiplies require 25 instruction cycles, and long ones require + * 45 instruction cycles. + * + * On return, overflow has occurred (%o1 is not zero) if and only if + * the Z condition code is clear, allowing, e.g., the following: + * + *	call	.umul + *	nop + *	bnz	overflow	(or tnz) + */ + +ENTRY(.umul) +	or	%o0, %o1, %o4 +	mov	%o0, %y			! multiplier -> Y +	andncc	%o4, 0xfff, %g0		! test bits 12..31 of *both* args +	be	LOC(mul_shortway)	! if zero, can do it the short way +	 andcc	%g0, %g0, %o4		! zero the partial product; clear N & V + +	/* +	 * Long multiply.  32 steps, followed by a final shift step. +	 */ +	mulscc	%o4, %o1, %o4	! 1 +	mulscc	%o4, %o1, %o4	! 2 +	mulscc	%o4, %o1, %o4	! 3 +	mulscc	%o4, %o1, %o4	! 4 +	mulscc	%o4, %o1, %o4	! 5 +	mulscc	%o4, %o1, %o4	! 6 +	mulscc	%o4, %o1, %o4	! 7 +	mulscc	%o4, %o1, %o4	! 8 +	mulscc	%o4, %o1, %o4	! 9 +	mulscc	%o4, %o1, %o4	! 10 +	mulscc	%o4, %o1, %o4	! 11 +	mulscc	%o4, %o1, %o4	! 12 +	mulscc	%o4, %o1, %o4	! 13 +	mulscc	%o4, %o1, %o4	! 14 +	mulscc	%o4, %o1, %o4	! 15 +	mulscc	%o4, %o1, %o4	! 16 +	mulscc	%o4, %o1, %o4	! 17 +	mulscc	%o4, %o1, %o4	! 18 +	mulscc	%o4, %o1, %o4	! 19 +	mulscc	%o4, %o1, %o4	! 20 +	mulscc	%o4, %o1, %o4	! 21 +	mulscc	%o4, %o1, %o4	! 22 +	mulscc	%o4, %o1, %o4	! 23 +	mulscc	%o4, %o1, %o4	! 24 +	mulscc	%o4, %o1, %o4	! 25 +	mulscc	%o4, %o1, %o4	! 26 +	mulscc	%o4, %o1, %o4	! 27 +	mulscc	%o4, %o1, %o4	! 28 +	mulscc	%o4, %o1, %o4	! 29 +	mulscc	%o4, %o1, %o4	! 30 +	mulscc	%o4, %o1, %o4	! 31 +	mulscc	%o4, %o1, %o4	! 32 +	mulscc	%o4, %g0, %o4	! final shift + +	/* +	 * Normally, with the shift-and-add approach, if both numbers are +	 * positive you get the correct result.  With 32-bit two's-complement +	 * numbers, -x is represented as +	 * +	 *		  x		    32 +	 *	( 2  -  ------ ) mod 2  *  2 +	 *		   32 +	 *		  2 +	 * +	 * (the `mod 2' subtracts 1 from 1.bbbb).  To avoid lots of 2^32s, +	 * we can treat this as if the radix point were just to the left +	 * of the sign bit (multiply by 2^32), and get +	 * +	 *	-x  =  (2 - x) mod 2 +	 * +	 * Then, ignoring the `mod 2's for convenience: +	 * +	 *   x *  y	= xy +	 *  -x *  y	= 2y - xy +	 *   x * -y	= 2x - xy +	 *  -x * -y	= 4 - 2x - 2y + xy +	 * +	 * For signed multiplies, we subtract (x << 32) from the partial +	 * product to fix this problem for negative multipliers (see mul.s). +	 * Because of the way the shift into the partial product is calculated +	 * (N xor V), this term is automatically removed for the multiplicand, +	 * so we don't have to adjust. +	 * +	 * But for unsigned multiplies, the high order bit wasn't a sign bit, +	 * and the correction is wrong.  So for unsigned multiplies where the +	 * high order bit is one, we end up with xy - (y << 32).  To fix it +	 * we add y << 32. +	 */ +#if 0 +	tst	%o1 +	bl,a	1f		! if %o1 < 0 (high order bit = 1), +	 add	%o4, %o0, %o4	! %o4 += %o0 (add y to upper half) +1:	rd	%y, %o0		! get lower half of product +	retl +	 addcc	%o4, %g0, %o1	! put upper half in place and set Z for %o1==0 +#else +	/* Faster code from tege@sics.se.  */ +	sra	%o1, 31, %o2	! make mask from sign bit +	and	%o0, %o2, %o2	! %o2 = 0 or %o0, depending on sign of %o1 +	rd	%y, %o0		! get lower half of product +	retl +	 addcc	%o4, %o2, %o1	! add compensation and put upper half in place +#endif + +LOC(mul_shortway): +	/* +	 * Short multiply.  12 steps, followed by a final shift step. +	 * The resulting bits are off by 12 and (32-12) = 20 bit positions, +	 * but there is no problem with %o0 being negative (unlike above), +	 * and overflow is impossible (the answer is at most 24 bits long). +	 */ +	mulscc	%o4, %o1, %o4	! 1 +	mulscc	%o4, %o1, %o4	! 2 +	mulscc	%o4, %o1, %o4	! 3 +	mulscc	%o4, %o1, %o4	! 4 +	mulscc	%o4, %o1, %o4	! 5 +	mulscc	%o4, %o1, %o4	! 6 +	mulscc	%o4, %o1, %o4	! 7 +	mulscc	%o4, %o1, %o4	! 8 +	mulscc	%o4, %o1, %o4	! 9 +	mulscc	%o4, %o1, %o4	! 10 +	mulscc	%o4, %o1, %o4	! 11 +	mulscc	%o4, %o1, %o4	! 12 +	mulscc	%o4, %g0, %o4	! final shift + +	/* +	 * %o4 has 20 of the bits that should be in the result; %y has +	 * the bottom 12 (as %y's top 12).  That is: +	 * +	 *	  %o4		    %y +	 * +----------------+----------------+ +	 * | -12- |   -20-  | -12- |   -20-  | +	 * +------(---------+------)---------+ +	 *	   -----result----- +	 * +	 * The 12 bits of %o4 left of the `result' area are all zero; +	 * in fact, all top 20 bits of %o4 are zero. +	 */ + +	rd	%y, %o5 +	sll	%o4, 12, %o0	! shift middle bits left 12 +	srl	%o5, 20, %o5	! shift low bits right 20 +	or	%o5, %o0, %o0 +	retl +	 addcc	%g0, %g0, %o1	! %o1 = zero, and set Z + +END(.umul)  | 
