1. Computer Science 210 Computer Organization
Type Conversions and Numeric I/O

2. Numeric Input and Output
# In Python 2.7
yourAge = input("Enter your age: ")
print yourAge
# In Python 3.2
yourAge = int(input("Enter your age: "))
print(yourAge)
In Python 3
there is no raw_input
input returns a string, which must be cast to an int
print automatically converts its argument to a string

3. Input and Output in LC-3 In LC-3, all I/O is character-based
Strings may be output (trap service routine PUTS) and input (our own subroutine)
Add subroutines GETS, INT, and TOSTRING to handle numeric I/O

4. The Numeric I/O Interface
Subroutine
Input Parameters
Output Parameters
PUTS
R0 - address of string
GETS
R1 - address of string
R2 – maximum size
(Upon RETurn, address starting at R1 will contain our string)
INT
R2 - integer value
TOSTRING
PUTS is built into the assembler (we get it for free)
Routines in red we must write ourselves
We stay away from R0, which is used by the LC-3 trap routines and by our own stack routines

5. Simple Raw Input (basis of GETS)
;; Author: Ken Lambert
;; This program prompts the user for a string, inputs it, and prints it.
.ORIG x3000
;; Register usage:
; R0 = the address of a string for output
; R1 = the address of a string for input
; R6 = the stack pointer
; Main program code
JSR INITSTACK ; Initialize the stack
LEA R0, PROMPT ; Display the prompt
PUTS
LEA R1, BUFFER ; Input a string
LD R2, MAXSIZE
JSR GETS
ADD R0, R1, #0 ; Output the string (same as LEA R0, BUFFER)
HALT
; Main program data variables
PROMPT .STRINGZ "Enter a string: ” ; The input prompt
BUFFER .BLKW ; String buffer for I/O (including null)
MAXSIZE .FILL ; Maximum number of characters

6. Print the Sum of Two Input Integers
; Main program code
JSR INITSTACK ; Initialize the stack
LEA R0, PROMPT ; Prompt for an integer
PUTS
LEA R1, BUFFER ; Input a string
LD R2, MAXSIZE
JSR GETS
JSR INT ; Convert to an integer and save it
ST R2, FIRST
ST R2, SECOND
LD R1, FIRST ; Add the two integers
LD R2, SECOND
ADD R2, R1, R2
LEA R1, BUFFER ; Convert the sum to a string and print it
JSR TOSTRING
ADD R0, R1, #0
HALT
; Main program data variables
PROMPT .STRINGZ "Enter an integer: " ; The input prompt
FIRST .BLKW ; Two integers
SECOND .BLKW 1
BUFFER .BLKW ; String buffer for I/O (including null)
MAXSIZE .BLKW ; Maximum number of characters

7. Convert a String of Digits to an int
def int(string):
number = 0
for digit in string:
number = 10 * number + ord(digit) - ord('0')
return number
The ord function returns a character’s integer ASCII value
int(digit) == ord(digit) - ord('0')

8. Code for subroutine INT
; Input parameter: R1 - the address of the string buffer
; Output parameter: R2 - the integer represented by the string
; Register usage:
; R3 = temporary working storage
; R4 = the pointer into the string buffer
INT PUSH R7, R1, R3, and R ; Save registers
AND R3, R3, #0 ; Clear the sum
ST R3, INTSUM
ADD R4, R1, #0 ; Set the pointer into the buffer
INTLOOP LDR R1, R4, #0 ; Get the next digit from the buffer
BRz ENDINT ; Quit when it's null
LD R2, ORDZERO ; Convert the digit to an int
ADD R1, R1, R2
ST R1, INTDIGIT
LD R1, INTSUM ; Multiply the sum by 10
LD R2, INT10
JSR MUL
LD R1, INTDIGIT ; Add int value of digit to the sum
ADD R3, R3, R1
ADD R4, R4, #1 ; Advance to the next character in the buffer
BR INTLOOP
ENDINT LD R2, INTSUM ; Set the output parameter
POP R4, R3, R1, and R7 ; Restore registers
RET
; Subroutine INT data variables
INTDIGIT .BLKW ; Holds the integer value of each digit
ORDZERO .FILL # ; ASCII for the digit '0' (negated)
INT10 .FILL #10 ; Base of 10
INTSUM .FILL #0 ; Holds the running total for the sum
Code for subroutine INT

9. Convert an int to a String of Digits
def str(number):
if number == 0: return '0'
string = ''
while number > 0:
remainder = number % 10
number //= 10
string = chr(remainder + ord('0')) + string
return string
The chr function returns the character corresponding to an integer ASCII value
str(singleDigitInt) == chr(singleDigitInt + ord('0'))

10. Convert an int to a String of Digits
def str(number):
if number == 0: return '0'
string = ''
while number > 0:
remainder = number % 10
number //= 10
string = chr(remainder + ord('0')) + string
return string
Prepending characters to a string is not easy when you’re representing the string as an array of memory cells

11. Convert an int to a String of Digits
def str(number):
if number == 0: return '0'
stack = Stack()
while number > 0:
remainder = number % 10
number //= 10
stack.push(chr(remainder + ord('0')))
string = ''
while not stack.isEmpty():
string += stack.pop()
return string
Use a stack to unload digits to the string in the reverse order

12. Code for subroutine TOSTRING
; Converts the integer in R2 to a string with base address R1
; Input parameters: R1 = base address of string buffer
; R2 = an integer
; Output parameter: R1 = base address of the string buffer
TOSTRING ST R1, STRBASE ; Save return address, parameters,
PUSH R7, R4, R3, and R2 ; and temporaries
AND R0, R0, #0 ; Push the null character onto the stack
JSR PUSH
ADD R1, R2, #0 ; Set the initial dividend
STRLOOP LD R2, INT10 ; Divide the number by 10
JSR DIV
LD R2, CHRZERO ; Convert the remainder and push it
ADD R0, R4, R2 ; onto the stack
ADD R1, R3, #0 ; Set the dividend to the quotient
BRp STRLOOP ; While quotient > 0
LD R1, STRBASE ; Move characters from the stack to the
POPLOOP JSR POP ; string buffer, stopping after the null
STR R0, R1, #0 ; character is moved
ADD R0, R0, #0
BRz ENDPOP
ADD R1, R1, #1
BR POPLOOP
ENDPOP LD R1, STRBASE ; Restore all registers
POP R2, R3, R4, and R7
RET
; Subroutine TOSTRING data variables (also uses INT10 from INT)
CHRZERO .FILL #48 ; ASCII for the digit '0'
STRBASE .BLKW 1 ; Base address of string buffer for the sum
Code for subroutine TOSTRING

13. Managing Dynamic Memory With a System Heap
For Monday
Managing Dynamic Memory
With a System Heap