Return with BCDE=0 if A=0.

0A77 47 FAsInteger MOV B,A

0A78 4F MOV C,A

0A79 57 MOV D,A

0A7A 5F MOV E,A

0A7B B7 ORA A

0A7C C8 RZ

Preserve HL and copy FACCUM to BCDE

0A7D E5 PUSH H

0A7E CD1D0A CALL FCopyToBCDE

Unpack FACCUM's mantissa to get the hidden most-significant-bit (see top of page) and get the sign back via an XOR to undo what FUnpackMantissas did with it. Preserve the sign in H

0A81 CD370A CALL FUnpackMantissas

0A84 AE XRA M Get sign back

0A85 67 MOV H,A

If FACCUM was negative then decrement the mantissa (two's complement?)

0A86 FC9B0A CM FMantissaDec

Shift mantissa in CDE right by (24-B) places. This gets the integer part of FACCUM into CDE, which is the whole point of the function.

0A89 3E98 MVI A,98

Shift mantissa right

0A8B 90 SUB B by (24-exponent) places?

0A8C CDC908 CALL FMantissaRtOnce WHY?

If floating point sign is negative (ie bit 7 of H set) then two's complement CDE to get the signed integer. We do two's complement by first adding 1, then negating.

0A8F 7C MOV A,H

0A90 17 RAL

0A91 DC9A08 CC FMantissaInc

0A94 0600 MVI B,00 Needed for FNegateInt.

0A96 DCB508 CC FNegateInt

Restore HL and return with the integer result in CDE.

0A99 E1 POP H

0A9A C9 RET

0A9B 1B FMantissaDec DCX D DE--

0A9C 7A MOV A,D If DE!=0xFFFF...

0A9D A3 ANA E

0A9E 3C INR A

0A9F C0 RNZ ... then return

0AA0 0D DCR C C--

0AA1 C9 RET

If FACCUM's exponent is >= 2^24, then it's too big to hold any fractional part - it is already an integer, so we just return.

0AA2 217201 Int LXI H,FACCUM+3

0AA5 7E MOV A,M

0AA6 FE98 CPI 98

0AA8 D0 RNC

Convert FACCUM to an integer in CDE. On returning, HL points to FACCUM's exponent byte (ie FACCUM+3).

0AA9 CD770A CALL FAsInteger

Now we need to convert the integer in CDE to a proper floating point value. To do this we first set FACCUM's exponent to 2^24.

0AAC 3698 MVI M,98

0AAE 79 MOV A,C

0AAF 17 RAL

0AB0 C35B08 JMP 085B