CS536 Homework 7 Solution

Question 1 | Question 2


Question 1:

Part a:

An expression that cannot be evaluated using just 2 registers, but can be evaluated using 3 registers:

expression AST pseudo code
 (a - b) - (c - d) 
      -
    /   \
   -     -
  / \   / \
 a   b c   d 
  load a into R0 
  load b into R1 
  R0 = R0 - R1
  load c into R1 
  load b into R2 
  R1 = R1 - R2
  R0 = R0 - R1

Part b:

An expression that cannot be evaluated using just 3 registers,but can be evaluated using 4 registers:

expression AST pseudo code
 ((a - b) - (c - d)) - ((e - f) - (g - h)) 
            -
         /     \
        /       \
       /         \
      -           -	
    /   \       /   \
   -     -     -     -
  / \   / \   / \   / \
 a   b c   d e   f g   h 
  load a into R0 
  load b into R1 
  R0 = R0 - R1
  load c into R1 
  load d into R2 
  R1 = R1 - R2
  R0 = R0 - R1
  load e into R1
  load f into R2
  R1 = R1 - R2
  load g into R2
  load h into R3
  R2 = R2 - R3
  R1 = R1 - R2
  R0 = R0 - R1

Question 2:

int numRegisters(ASTNode node) {

    if isLeaf(node)
        return 1
        
    else
        N1 = numRegisters(node.leftChild)
        N2 = numRegisters(node.rightChild)
        
       if (N1 == N2) 
           return N1 + 1
       else 
           return max(N1, N2)
}

Last Updated: 4/11/2024     © 2024 CS 536