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-/*
- * 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)
- */
-
-#include "DEFS.h"
-FUNC(.umul)
- or %o0, %o1, %o4
- mov %o0, %y ! multiplier -> Y
- andncc %o4, 0xfff, %g0 ! test bits 12..31 of *both* args
- be Lmul_shortway ! if zero, can do it the short way
- andcc %g0, %g0, %o4 ! zero the partial product and clear N and 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
-
-Lmul_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