HP HP-16C - Section 5: Floating-Point Numbers; Converting to Floating-Point Decimal Mode; Conversion in the Stack

54 Section 4: Arithmetic and Bit Manipulation Functions
Example: The calculation of (-88 = 11) in binary with word size 5
and 2’s Complement mode is shown below.
X
s,
- 8/0 X 11000, Five-bit result in X.
_ e
/1| -88 01011111101 01000 Ten-bit representation of
——— ——— -88,(split between Y- and
Y Z Z-registers.
Keystrokes Display ([STATUS]: 2-05-1000)
1000 01000 b Least significant bits of
10-bit dividend go into Z-
register.
11101 11101 b Most significant bits of
10-bit dividend go into Y-
register.
1011 (9](DBL=] 11000 b Quotient.
(9][cF]3 11000 b Restores suppression of
leading zeros.
Double Remainder
The function operates like except that the remainder is
returned instead of the quotient. If the quotient exceeds 64 bits,
Error O results.
The remainder is determined as for the function (page 43),
with the sign of the result matching the sign of the dividend.
Example: Applying Double Divide
5714AF2,
7TE14684,4
Although the result is a fraction, this problem can be solved in
Integer mode by first finding the integer quotient of
Compute the quotient of to 16 hexadecimal places.
16 zeros
———
5714AF2000...0,4
7TE14684
and then placing a decimal point to the left of the result (thereby
dividing the result by 25%). Use to accommodate a numerator
this large.