# .sect .text; .sect .rom; .sect .data; .sect .bss /* Multiplies two double-precision floats, then splits the product into * fraction and integer, both as floats, like modf(3) in C, * http://en.cppreference.com/w/c/numeric/math/modf * * Stack: ( a b -- fraction integer ) */ .sect .text .define .fif8 .fif8: ldc1 f0, 8(sp) ! f0 = a ldc1 f2, 0(sp) ! f2 = b mul.d f0, f0, f2 ! f0 = a * b abs.d f2, f0 ! f2 = abs(f0) li at, ha16[max_power_of_two] ldc1 f4, lo16[max_power_of_two] (at) ! f4 = max power of two mov.d f6, f2 ! we're going to assemble the integer part in f6 c.lt.d 0, f4, f2 ! if absolute value too big, it must be integral bc1t 0, return nop ! Crudely strip off the fractional part. add.d f6, f2, f4 ! f6 = absolute value + max power of two sub.d f6, f6, f4 ! f6 -= max_power_of_two ! The above might round, so correct that. li at, ha16[one] ldc1 f8, lo16[one] (at) ! f8 = 1.0 1: c.le.d 0, f6, f2 ! if result <= absolute value, stop bc1t 0, 2f nop sub.d f6, f6, f8 ! result -= 1.0 b 1b nop 2: ! Correct the sign of the result. mtc1 zero, f8 mthc1 zero, f8 ! f8 = 0.0 c.lt.d 0, f0, f8 ! if original value was negative bc1f 0, 1f nop neg.d f6, f6 ! negate the result 1: return: sdc1 f6, 0(sp) ! store integer part sub.d f6, f0, f6 ! calculate fractional part sdc1 f6, 8(sp) ! store fractional part jr ra nop ! doubles >= MAXPOWTWO are already integers .sect .rom max_power_of_two: .dataf8 4.503599627370496000E+15 one: .dataf8 1.0