From 46b06e9c1c7d57ae0008b3a282110d8ce4345b5f Mon Sep 17 00:00:00 2001 From: Mike Frysinger Date: Thu, 25 Aug 2005 23:50:55 +0000 Subject: import different optimized versions of div funcs based upon target sparc arch --- libc/sysdeps/linux/sparc/sparcv7/umul.S | 153 ++++++++++++++++++++++++++++++++ 1 file changed, 153 insertions(+) create mode 100644 libc/sysdeps/linux/sparc/sparcv7/umul.S (limited to 'libc/sysdeps/linux/sparc/sparcv7/umul.S') 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) -- cgit v1.2.3