2018-09-10 21:20:52 +00:00
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#
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.sect .text; .sect .rom; .sect .data; .sect .bss
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/* Multiplies two double-precision floats, then splits the product into
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* fraction and integer, both as floats, like modf(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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*/
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.sect .text
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.define .fif8
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.fif8:
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ldc1 f0, 8(sp) ! f0 = a
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ldc1 f2, 0(sp) ! f2 = b
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mul.d f0, f0, f2 ! f0 = a * b
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abs.d f2, f0 ! f2 = abs(f0)
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2018-09-12 20:28:40 +00:00
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lui at, ha16[max_power_of_two]
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2018-09-10 21:20:52 +00:00
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ldc1 f4, lo16[max_power_of_two] (at) ! f4 = max power of two
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mov.d f6, f2 ! we're going to assemble the integer part in f6
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c.lt.d 0, f4, f2 ! if absolute value too big, it must be integral
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bc1t 0, return
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nop
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! Crudely strip off the fractional part.
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add.d f6, f2, f4 ! f6 = absolute value + max power of two
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sub.d f6, f6, f4 ! f6 -= max_power_of_two
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! The above might round, so correct that.
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2018-09-12 20:28:40 +00:00
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lui at, ha16[one]
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2018-09-10 21:20:52 +00:00
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ldc1 f8, lo16[one] (at) ! f8 = 1.0
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1:
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c.le.d 0, f6, f2 ! if result <= absolute value, stop
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bc1t 0, 2f
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nop
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sub.d f6, f6, f8 ! result -= 1.0
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b 1b
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nop
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2:
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! Correct the sign of the result.
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mtc1 zero, f8
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mthc1 zero, f8 ! f8 = 0.0
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c.lt.d 0, f0, f8 ! if original value was negative
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bc1f 0, 1f
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nop
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neg.d f6, f6 ! negate the result
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1:
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return:
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sdc1 f6, 0(sp) ! store integer part
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sub.d f6, f0, f6 ! calculate fractional part
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sdc1 f6, 8(sp) ! store fractional part
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jr ra
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nop
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! doubles >= MAXPOWTWO are already integers
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.sect .rom
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max_power_of_two:
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.dataf8 4.503599627370496000E+15
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one:
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.dataf8 1.0
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