ROM subroutines: convert INT64 to DOUBLE

ROM subroutines

Convert INT64 to DOUBLE

This subroutine has two main branches:

when intreger argument has more than 53 bits (including the most significant one that will be hidden!), we must shift the value right to get normalized mantissa;
othewise we'll shift argument in opposite direction until 1 will appear in the 52th - the most significant in mantissa - bit (bits 0-51 will form normalized mantissa at this moment).

In both cases the process ends when bit number 52 will be 1, and bits 53-63 are 0 - the character feature of normalized mantissa. 52th bit is not stored - so called hidden bit.

When INT64 number is equal to 2⁵²₁₀ = 2³⁴₁₆, its mantissa will be normalized promptly. Hexadecimal exponent in this case can be calculated as 34 + 3FF = 433 (this constant is used in the second insruction of the subroutine).

addr code mnemonic action comments
EE0...1A4 4271 0017 BZ $71, 0017 branch to 200 if $71 = 0 exit if argument is 0
1A8 E072 4330 SETH $72, 4330 $72 <== 4330..00 this value has exponent when INT64 already forms normal mantissa (53 bits)
1AC 3173 7100 CMPI $73, $71, 00 compare $71 with 0 and set $73 is argument negative?
1B0 4873 0003 BNN $73, 0003 branch to 1BC if $73 >= 0 bypass if sign is plus
1B4 C571 7100 NORI $71, $71, 00 $71 <== NOT $71 convert to positive value
1B8 E771 0001 INCL $71, 0001 $71 <== $71 + 1
1BC E074 FFE0 SETH $74, FFE0 $74 <== FFE0..00 constant to test if INT64 has more bits than mantissa
1C0 E075 0010 SETH $75, 0010 $75 <== 0010..00 set the less signisicant exponent's bit
1C4 C876 7471 AND $76, $74, $71 $76 <== $74 AND $71 reset 53 lower bits of argument
1C8 4276 0004 BZ $76, 0004 branch to 01D8 if $76 = 0 further if higher bits are empty
1CC 3D71 7101 SRI $71, $71, 01 right shift $71 INT64 must be shifted right to form normal mantissa
1D0 2072 7275 ADD $72, $72, $75 $72 <== $72 + $75 INC exponent
1D4 F1FF FFFC JMPB FFFFFC branch to 01C4 repeat the cycle
1D8 E074 FFF0 SETH $74, FFF0 $74 <== FFF0..00 constant to reset mantissa except its 52th bit
1DC C876 7471 AND $76, $74, $71 $76 <== $74 AND $71 clear mantissa to test the most significant bit
1E0 4A76 0004 BNZ $76, 0004 branch to 1F0 finish cycle when it is 1
1E4 3971 7101 SLI $71, $71, 01 left shift $71 INT64 must be shifted left to form normal mantissa
1E8 2472 7275 SUB $72, $72, $75 $72 <== $72 - $75 DEC exponent
1EC F1FF FFFC JMPB FFFFFC branch to 1DC repeat the cycle
1F0 2471 7175 SUB $71, $71, $75 $71 <== $71 - $75 clear the most sidnificant (hidden) bit in mantissa
1F4 C071 7172 OR $71, $71, $72 $71 <== $71 OR $72 combine mantissa with exponent
1F8 4873 0002 BNN $73, 0002 branch to 200 if $73 >= 0 bypass if argument was not negative
1FC E471 8000 INCH $71, 8000 $71 <== $71 + 80..00 set negative sign
200 9F6F 6F00 GOI $6F, $6F, 0 return using $6F (finish subroutine)

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E-MMI software