Question
LC-3 ASSEMBLY LANGUAGE
CODE TEMPLATE:
.ORIG x3000 ; Program begins here
;-------------
;Instructions
;-------------
LD R6, Convert_addr ; R6 <-- Address pointer for Convert
LDR R1, R6, #0 ; R1 <-- VARIABLE Convert
;-------------------------------
;ENTER CODE HERE
;--------------------------------
HALT
;---------------
;Data
;---------------
Convert_addr .FILL xD000 ; The address of where to find the data
.ORIG xD000 ; Remote data
Convert .FILL xABCD ; <----!!!NUMBER TO BE CONVERTED TO BINARY!!!
;---------------
;END of PROGRAM
;---------------
.END
Objective The purpose of this assignment is to give students practice with Lyo, left-shifting, multiplying by 2, and useful 2's complement logic. High Level Description Load any valid number into a register from the memory address specified in your assn 3template and output it to the console as 16-bit two's complement binary (i.e. the native format of the LC-3) Note: Valid numbers are (#-32768, #327671 (decima or [x0000, xFFFF] (hex) Your Tasks You do not yet know how to take a multi-digit decimal number from user input and convert it to binary, so for this assignment you are going to get the assembler to do that part for you: you will use the .FILL pseudo-op to take a literal (decimal or hex, as you wish) and translate it into 16-bit two's comp. binary, and store that value in the indicated memory location; and then you will Load that value from memory into R1 You MUST use the provided assn3.asm template to set this up: it ensures that the number to be converted is always stored in the same location (the memory address specified in your template) so we can test your work; make sure you fully understand the code we provide At this point, your value will be stored in R1: it is now your job to extract the 1's and 0 sfrom the number and print them out to the console one by one, from left to right. Important things to consider: Recall the difference between a positive number and a negative number in 2's complement binary: if the most significant bit (MSB) is 0, the number is positive; if it is 1, the number is negative The BR instruction has parameters (n, z, p) which tell it to check whether a value is anch negative, zero, or positive (or any combination thereof). Once you are done inspecting the MSB, how would you shift the next bit into its place so you could perform the next iteration? (hint: the answer is in the objectives) Pseudocode: for(i 5 downto 0) if bit[i] is a 0) print a 0 else print a 1 shift left
Explanation / Answer
public category NonRecursiveTraversal non-public TreeNode root;
personal static category TreeNode
}
public void createBinaryTree (Integer[] arr) a pair of * i + 1;
final int right = a pair of * i + 2;
if (arr[left] != null)
if (right < arr.length && arr[right] != null)
}
}
}
public void preOrder() whereas (!stack.empty()) whereas (!stack.empty()) whereas (!stack.isEmpty() || node != null) else whereas (!stack.isEmpty() || node != null) whereas (node != null)
node = stack.pop();
// (node.right) != null suggests that this node is parent of that little subtree
// stack.pop != null ensures it had been not the foundation.
if (node.right != null && !stack.isEmpty() && node.right == stack.peek()) else
