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%%HP: T(3)A(R)F(.);
\<< 0 0 \-> A B AX BX
\<< A DIG 3 + DUP 3 + SUB OBJ\-> 'AX' STO B DIG 3 + DUP 3 + SUB OBJ\-> 'BX' STO AX BX - ABS DIG >
IF
THEN AX BX >
IF
THEN A
ELSE B
END KILL
END A 1 DIG 1 + SUB OBJ\-> B 1 DIG 1 + SUB OBJ\->
AX BX - DUP 0 \>=
IF
THEN 10 SWAP ^ ROT *
ELSE NEG 10 SWAP ^ *
END DUP SIZE UNROT

This is just a header. It's added (and used) by the transfer program. You wouldn't add that when you write the program.

Loads the two numbers on the stack as local variables A and B. Prepares local variables AX and BX, which will be used to store the numbers' exponents.

Extracts the exponents from the number format and stores them.

If one number is so much larger than the other, then there is no point in adding them. If the difference in exponents is larger than the number of digits used, then this program just returns the larger (in absolute value) of the two numbers and terminates.

Extracts the mantissas from the numbers.

To add two numbers, the decimal point has to be lined up correctly. This part of the program does that, then leaves the larger number (absolutely) on level 1 of the stack.

Gets the number of digits in the larger number, which will be needed to determine the exponent of the final sum. Then gets it out of the way, so the two numbers to be added are on levels 1 and 2 of the stack.

+ DUP SIZE UNROT DUP DUP 0 ==
IF
THEN DROP DROP DROP DROP ZERO KILL
END 0 >
IF
THEN "+" SWAP \->STR +
ELSE \->STR
END UNROT - \->Q AX BX MAX + DUP 0 \>=
IF
THEN "E+"
ELSE "E-"
END SWAP ABS DUP \->STR

Adds the two numbers. Yup, that's it, right there. The advantage of using the HP50g is that it makes this step that simple.

Gets the size of the sum, which will be later used for determining the exponent of the sum. Then it gets it out of the way, leaving the sum on level 1 of the stack.

Gets ready for two comparisons. The first comparison is to zero. If it's zero, it just drops the second comparison, the sum, and the two digits (four drops total), puts ZERO on the stack, and terminates.

Determines if the number is bigger or smaller than zero, and puts the appropriate sign at the beginning.

Subtracts the digits of the sum, and the larger of the two summands. The SIZE command results in an approximate number, so it has to be converted to an integer with the ->Q command. Adds this to the larger of the original exponents to determine the exponent of the sum.

Determines the sign of the exponent, and puts it in the proper format.

Converts the exponent to a string.

SWAP SIZE DUP 1 \<=
IF
THEN DROP "00" SWAP +
ELSE 2 ==
IF
THEN "0" SWAP +
END
END DUP2 OBJ\-> SWAP "1" + 2 3 SUB OBJ\-> * DUP 999 > SWAP -999 < DUP UNROT OR
IF
THEN
IF
THEN ZERO KILL
ELSE DROP DROP INF SWAP 1 1 SUB "1" + OBJ\-> -1 ==
IF
THEN NGTV
END KILL
END
ELSE DROP
END + SWAP 1 DIG 1 + SUB SWAP +
\>>



\>>

Puts the numerical exponent back in position 1 and checks how big it is. I probably should have done this with a CASE command rather than IF. Oh, well. Makes sure the exponent has leading zeros if necessary.

I'm thinking maybe this should have been done in a different order. Oh, well, again. Converts the exponent back into a number. Checks if it's bigger than +999 or smaller than -999. If it's the former, puts zero on the stack and terminates. HOWEVER, this is a bug. I should have had three DROPs, then ZERO, then KILL.

If it was the latter, it puts INF on the stack, and checks if it should be positive or negative. I believe this part does not have the bug.

This last condition gets rid of a truth value that was used in the second IF statement.

Combines the exponent with its sign, I think.

Takes the mantissa and crops off any excess digits.

Finally puts everything in the right format.