66f93f08c5
When I wrote fef 8, I forgot to test denormalized numbers. Oops. Now fix two of my mistakes: - When checking for zero, `extrwi r6, r3, 22, 12` needs to be `extrwi r6, r3, 20, 12`. There are only 20 bits to extract. - After the multiplication by 2**64, I forgot to put the fraction in [0.5, 1) or (-1, 0.5] by setting IEEE exponent = 1022. Teach fif 8 about signed zero and NaN. In ncg/table, change cmf so NaN is not equal to any value, and comment why ordered comparisons don't work with NaN. Also add cost for fctwiz, remove extra `uses REG`. Edit comment in cfu8.s because the conditional branch might be before or after fctwiz.
64 lines
1.8 KiB
ArmAsm
64 lines
1.8 KiB
ArmAsm
.sect .text
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! Multiplies two single-precision floats, then splits the product into
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! fraction and integer, both as floats, like modff(3) in C,
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! http://en.cppreference.com/w/c/numeric/math/modf
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!
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! Stack: ( a b -- fraction integer )
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.define .fif4
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.fif4:
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lfs f1, 4(sp)
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lfs f2, 0(sp)
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fmuls f1, f1, f2 ! f1 = a * b
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stfs f1, 0(sp)
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lwz r3, 0(sp) ! r3 = word of float bits
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! IEEE single = sign * 1.fraction * 2**(exponent - 127)
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! sign exponent fraction
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! 0 1..8 9..31
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!
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! Subtract 127 from the IEEE exponent. If the result is from
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! 0 to 23, then the IEEE fraction has that many integer bits.
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extrwi r5, r3, 8, 1 ! r5 = IEEE exponent
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addic. r5, r5, -127 ! r5 = nr of integer bits
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blt 3f ! branch if no integer
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cmpwi r5, 24
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bge 4f ! branch if no fraction
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! fall through if integer with fraction
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! f1 has r5 = 0 to 23 integer bits in the IEEE fraction.
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! There are 23 - r5 fraction bits.
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li r6, 23
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subf r6, r5, r6
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srw r3, r3, r6
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slw r3, r3, r6 ! clear fraction in word
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! fall through
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1: stw r3, 0(sp)
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lfs f2, 0(sp) ! integer = high word, low word
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fsubs f1, f1, f2 ! fraction = value - integer
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2: stfs f1, 4(sp) ! push fraction
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stfs f2, 0(sp) ! push integer
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blr
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! f1 is a fraction without integer (or zero).
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! Then integer is zero with same sign.
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3: extlwi r3, r3, 1, 0 ! extract sign bit
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stfs f1, 4(sp) ! push fraction
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stw r3, 0(sp) ! push integer = zero with sign
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blr
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! f1 is an integer without fraction (or infinity or NaN).
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! Unless NaN, then fraction is zero with same sign.
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4: fcmpu cr0, f1, f1
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bun cr0, 5f
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extlwi r3, r3, 1, 0 ! extract sign bit
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stw r3, 4(sp) ! push fraction = zero with sign
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stfs f1, 0(sp) ! push integer
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blr
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! f1 is NaN, so both fraction and integer are NaN.
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5: fmr f2, f1
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b 2b
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