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
INT
R2 - integer value
TOSTRING
We stay away from R0, which is used by the LC-3 trap routines and by our own stack routines

5. Simple Raw Input ;; 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 nul
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