Base conversion on HP-41

; BASE TO DECIMAL

01 LBL "BD"
02 XEQ 06 ; input base
03 AON
04 STOP ; input number

05 AOFF ; prepare for main loop
06 CLST ; X=0, Z=0, T=0

07 LBL 01 ; base 10 number will accumulate in Y register
08 >"+" ; append blanks in alpha until the first character
09 X<> O ; of the alpha input is pushed into alpha char
10 X=0? ; 15-21 (O register).

12 X<> M ; In the loop, the next char which appears in O is
13 R^ ; exchanged with M and the stack is rolled up to
14 X<> N ; preserve M in the stack. The N register is then
15 "&#92;&#48;0&#92;&#48;8" ; bought into X and O is stored in N. The alpha
16 X<> N ; register (MNO) now contains only the net character.
17 RDN ; 0x0008 is appended. N is then returned with its
18 X<> M ; original content, the stack is rolled down and then
19 E ; the M register is restored and our next character
20 * ; now appears as a decimal number in X. This number
21 39 ; is multiplied by 1 so it becomes normalized and then
22 - ; 39 is subtracted so we may test whether our char is
23 X>0? ; a digit 0-9 (row 3 in the next table) or a letter
24 DSE X ; (row 4). The X register is then transformed to its
25 9 ; true decimal value.
26 + ; It is then tested to see if it is a valid number
27 X<0? ; (a blank would yield a negative at this point
28 GTO 02 ; causing a jump to LBL 02).
29 X<>Y ; If not a blank, the accumulated result is multiplied
30 RCL 06 ; by the base and the next digit is added.
31 *
32 +
33 . ; Ensure the Y register is the only nonzero stack
34 GTO 01 ; register when the jump is made back to LBL 01.

35 LBL 02 ; End the routine by rolling down the stack to bring
36 RDN ; the Y register result into X and the alpha register
37 CLA ; is cleared.
38 RTN

; DECIMAL TO BASE

39 LBL "DB"
40 XEQ 06 ; input base
41 STOP ; input number

42 "\27 " ; single quote followed by 13 spaces
43 RCL M ; Recall 7 spaces into the stack. These blanks float
44 X<>Y ; up and down in the stack throughout the main loop
45 LBL 03 ; and are not used until line 74.

46 ENTER^ ; The main loop. At the start the remaining base 10
47 INT ; number is in X and the 7 spaces in Y. The base b
48 RCL 06 ; digits are built up starting with the least
49 MOD ; significant digits. The base 10 equivalent is
50 9 ; computed at lines 48-49. Lines 50-57 convert this
51 - ; decimal number to its alpha equivalent stored in O.
52 X>0?
53 ISG X
54 LBL 04
55 39
56 +
57 10^X
58 STO O ; Line 58 acts as an append to the remaining char in
59 >" " ; alpha. The next 4 lines do an alpha shift to
60 CLX ; prepare M and N for the next char (digit) to be
61 X<> O ; appended.
62 X<> N
63 STO M
64 RDN ; Calculate the remaining base 10 result and branch
65 RCL 06 ; to LBL 03 as long as this number is nonzero.
66 /
67 INT
68 X#0?
69 GTO 03

70 LBL 05 ; Pad space char in alpha so the result is left
71 >" " ; justified.
72 CLX
73 RCL M
74 X#Y? ; Comparison test with the 7 spaces.
75 GTO 05
76 CLX ; Alpha shift so that the final digits are in M
77 X<> O ; and N
78 X<> N
79 STO M
80 CLST ; Finish by clearing the stack view in the alpha
81 AVIEW
82 RTN

83 LBL 06
84 "BASE ?"
85 PROMPT
86 STO 06
87 "SOURCE ?"
88 AVIEW
89 RTN

90 END