Lines Matching refs:SO
31 # SOFTWARE so long as this entire notice is retained without alteration
633 # _real_ovfl() so that the operating system enabled overflow handler #
748 # overflow is enabled AND overflow, of course, occurred. so, we have the EXOP
792 # the src operand is definitely a NORM(!), so tag it as such
873 # _real_unfl() so that the operating system enabled overflow handler #
998 # overflow is enabled AND overflow, of course, occurred. so, we have the EXOP
1066 # the src operand is definitely a NORM(!), so tag it as such
1263 # so, since the emulation routines re-create them anyways, zero exception field
1294 beq.b fu_extract # yes, so it's monadic, too
1517 # so, since the emulation routines re-create them anyways, zero exception field.
1524 # the src can ONLY be a DENORM or an UNNORM! so, don't make any big subroutine
1591 # ("fmov.x fpm,-(a7)") if so,
1683 # for snan,operr,ovfl,unfl, src op is still in FP_SRC so just
1732 # three instruction exceptions don't update the stack pointer. so, if the
1819 # so, since the emulation routines re-create them anyways, zero exception field
1839 beq.b fu_extract_p # yes, so it's monadic, too
1911 # addressing mode was (a7)+. if so, we'll need to shift the
2076 # so, since the emulation routines re-create them anyways, zero exception field.
2142 # addressing mode was -(a7). if so, we'll need to shift the
2466 # So, we must check to see if it's disabled and handle that case separately.
2509 # so, now it's immediate data extended precision AND PACKED FORMAT!
2593 # now, we're left with ftst and fcmp. so, first let's tag them so that they don't
2663 # enabled? if so, then we have to stuff an overflow frame into the FPU.
2679 # the enabled exception was inexact. so, if it occurs with an overflow
2953 # The FPU is disabled and so we should really have taken the "Line
2954 # F Emulator" exception. So, here we create an 8-word stack frame
3143 # cause an operr so we don't need to check for them here.
3726 # through here. so can double and single precision.
3771 # be incorrect for some cases and need to be adjusted. So, this package #
3856 # exception. if so, branch directly to that handler's entry point.
3860 # check to see if the FPU is disabled. if so, jump to the OS entry
3867 # so, convert the F-Line exception stack frame to an FP Unimplemented
3889 # so, we need to convert the F-Line exception stack frame into an
3978 # FP instruction emulation. If so, then an FP exception of the correct #
3982 # Trace exception is enabled. If so, then we must create a Trace #
3987 # may flag that a BSUN exception should be taken. If so, then the #
4133 # into the machine. the frestore has already been executed...so, the fmov.l
4185 # often so that's why it gets lower priority.
4213 # the enabled exception was inexact. so, if it occurs with an overflow
4305 # the ftrapcc instruction should take a trap. so, here we must create a
4361 # so, the least significant WORD of the stacked effective address got
4363 # so that the rte will work correctly without destroying the result.
4463 # the fscc instruction should take a trace trap. so, here we must create a
5041 # k = N mod 4, so in particular, k = 0,1,2,or 3. #
5069 # k = N mod 4, so in particular, k = 0,1,2,or 3. #
5398 #--REGISTERS SAVED SO FAR: D0, A0, FP2.
5472 #--REGISTERS SAVED SO FAR: FP2.
5585 #--If compact form of abs(arg) in d0=$7ffeffff, argument is so large that
5634 #--FIND THE REMAINDER OF (R,r) W.R.T. 2**L * (PI/2). L IS SO CHOSEN
5743 # k = N mod 2, so in particular, k = 0 or 1. #
6016 #--If compact form of abs(arg) in d0=$7ffeffff, argument is so large that
6065 #--FIND THE REMAINDER OF (R,r) W.R.T. 2**L * (PI/2). L IS SO CHOSEN
6375 #--SO IF F IS CHOSEN TO BE CLOSE TO X AND ATAN(F) IS STORED IN
6852 # 2.3 Calculate J = N mod 64; so J = 0,1,2,..., #
6854 # 2.4 Calculate M = (N - J)/64; so N = 64M + J. #
6920 # rounded to 62 bits so that the last two bits of T are #
7009 # 2.2 Calculate J = N mod 64; so J = 0,1,2,..., #
7011 # 2.3 Calculate M = (N - J)/64; so N = 64M + J. #
7048 # rounded to 62 bits so that the last two bits of T are #
7658 # here, we build the result in a tmp location so as not to disturb the input
8126 # 1/F are also tabulated so that the division in (Y-F)/F #
8360 #--ALSO NOTE THAT THE VALUE 1/F IS STORED IN A TABLE SO THAT NO
9140 #--REGISTERS SAVED SO FAR ARE (IN ORDER) FPCR, D0, FP1, a1, AND FP2.
9174 #--REGISTERS SAVE SO FAR ARE FPCR AND D0
9237 #--REGISTERS SAVED SO FAR ARE (IN ORDER) FPCR, D0, FP1, a1, AND FP2.
9458 # answer is inexact, so set INEX2 and AINEX in the user's FPSR.
9660 # the src will force the dst to a DENORM value or worse. so, let's
9691 # The source input is below 1, so we check for denormalized numbers
10218 # so, normalize the mantissa, add 0x6000 to the new exponent,
10296 # dst so we don't screw up the version passed to us.
11559 # an exception. If so, return the default overflow/underflow result #
11881 # a normalized number that rounded down to a 2. so, redo the entire operation
12053 # if the result would have overflowed/underflowed. If so, use unf_res() #
12083 bne.w fin_not_ext # no, so go handle dbl or sgl
12086 # precision selected is extended. so...we cannot get an underflow
12087 # or overflow because of rounding to the correct precision. so...
12092 bset &neg_bit,FPSR_CC(%a6) # yes, so set 'N' ccode bit
12103 bne.w fin_not_ext # no, so go handle dbl or sgl
12108 bset &neg_bit,FPSR_CC(%a6) # yes, so set 'N' ccode bit
12303 # the move in MAY overflow. so...
12376 # an exception. If so, return the default overflow/underflow result #
12649 # or a normalized number that rounded down to a 1. so, redo the entire
12759 # So, determine the sign and return a new INF (w/ the j-bit cleared).
12828 # occurred. If so, return default underflow/overflow result. Else, #
12859 # precision selected is extended. so...we can not get an underflow
12860 # or overflow because of rounding to the correct precision. so...
13083 # the move in MAY underflow. so...
13231 # norms. Denorms are so low that the answer will either be a zero or a #
13279 # so, we could either set these manually or force the DENORM
13337 # norms. Denorms are so low that the answer will either be a zero or a #
13381 # so, we could either set these manually or force the DENORM
13449 # exponent would take an exception. If so, use unf_res() or ovf_res() #
13482 # precision selected is extended. so...we can not get an underflow
13483 # or overflow because of rounding to the correct precision. so...
13698 # the move in MAY underflow. so...
13851 # 'N' bit for a negative QNAN or SNAN input so we must squelch it here.
13912 # signs are the same, so must determine the answer ourselves.
13926 # signs are the same, so must determine the answer ourselves.
13961 # an exception. If so, return the default overflow/underflow result #
14176 # a normalized number that rounded down to a 2. so, redo the entire operation
14302 # an exception. If so, return the default overflow/underflow result #
14512 # or a normalized number that rounded down to a 1. so, redo the entire
14617 # addsub_scaler2() - scale the operands so they won't take exc #
14639 # occurred. If so, return default result and maybe EXOP. Else, insert #
14868 # ok, so now the result has a exponent equal to the smallest normalized
14874 # so, we do this be re-executing the add using RZ as the rounding mode and
14975 # the signs are the same. so determine whether they are positive or negative
15036 # ok, so it's not an OPERR. but, we do have to remember to return the
15070 # addsub_scaler2() - scale the operands so they won't take exc #
15092 # occurred. If so, return default result and maybe EXOP. Else, insert #
15321 # ok, so now the result has a exponent equal to the smallest normalized
15327 # so, we do this be re-executing the add using RZ as the rounding mode and
15428 # the signs are opposite, so, return a ZERO w/ the sign of the dst ZERO
15488 # ok, so it's not an OPERR. but we do have to remember to return
15536 # an exception. If so, return the default overflow/underflow result #
15670 bne.w fsqrt_sd_normal # yes, so no underflow
15780 # the move in MAY underflow. so...
15784 bne.w fsqrt_sd_ovfl # yes, so overflow
16024 # scale_sqrt(): scale the input operand exponent so a subsequent #
16283 # the displacement value to the stacked PC so that when an "rte" is #
16340 # counter. However, the true branch may set bsun so we check to see #
18057 # is enabled. if so, don't store result and correct stack frame
18186 # from an "Unimplemented Effective Address" exception handler. So, we #
18201 # so that the move can occur outside of this routine. This special #
18202 # case is required so that moves to the system stack are handled #
19054 clr.l %d2 # yes, so index = 0
19414 # mode so we don't even worry about this tricky case here : )
19448 # So, for -(an), we must subtract 8 off of the stacked <ea> value #
19771 # separate SNANs and DENORMs so they can be loaded w/ special care.
19788 fmov.s (%a0), %fp0 # no, so can load it regular
20203 # to-memory routines, then a special exit must be made so that the #
20373 # we copy the extended precision result to FP_SCR0 so that the reserved
20391 # the stack frame. so, leave it in FP_SRC for now and deal with it later...
20414 # so _mem_write2() handles this for us.
20500 # so, denorm and round and then use generic store single routine to
20552 # it's definitely an overflow so call ovf_res to get the correct answer
20571 smi %d1 # set if so
20607 # if exp now equals one, then it overflowed so call ovf_res.
20725 rts # no; so we're finished
20729 # so, denorm and round and then use generic store double routine to
20773 # it's definitely an overflow so call ovf_res to get the correct answer
20792 smi %d1 # set if so
20818 # if exp now equals one, then it overflowed so call ovf_res.
21037 # so _mem_write2() handles this for us.
21357 # increment by two. For any a7 update, set the mia7_flag so that if #
21421 so that if #
21629 # value in d0. The FP number can be DENORM or SNAN so we have to be #
21744 # all bit would have been shifted off during the denorm so simply
21775 # in memory so as to make the bitfield extraction for denormalization easier.
21991 # last operation done was an "and" of the bits shifted off so the condition
21992 # codes are already set so branch accordingly.
22000 # no bits were shifted off so don't set the sticky bit.
22009 # some bits were shifted off so set the sticky bit.
22161 roxr.w FTEMP_HI(%a0) # mant is 0 so restore v-bit
22162 roxr.w FTEMP_HI+2(%a0) # mant is 0 so restore v-bit
22182 roxr.w FTEMP_HI(%a0) # mant is 0 so restore v-bit
22183 roxr.w FTEMP_HI+2(%a0) # mant is 0 so restore v-bit
22236 # this function. so, as long as we don't disturb it, we are "returning" it.
22394 # hi(man) is all zeroes so see if any bits in lo(man) are set
22428 # exponent would go < 0, so only denormalize until exp = 0
22464 # whole mantissa is zero so this UNNORM is actually a zero
22944 # facc_in_x() - the fetch failed so jump to special exit code #
23056 # If so, set INEX1 in USER_FPSR. #
23082 mov.l 0x4(%a0),FP_SCR0_HI(%a6) # so we don't alter it
23242 # routine is necessary. If so, then check the sign of the exponent
23446 fdiv.x %fp1,%fp0 # exp is negative, so divide mant by exp
23449 fmul.x %fp1,%fp0 # exp is positive, so multiply by exp
23909 # to overflow. Only a negative iscale can cause this, so
23951 # so, we do the multiplication of the exponents and mantissas separately.
24109 # in extended precision, so the use of a previous power-of-ten
24256 bgt.b no_sft # if so, don't shift
24574 bra.b end_bstr # finished, so exit
24618 # So, we first call restore() which makes sure that any updated #
Indexes created Thu Oct 23 22:10:10 GMT 2025