CS536 Homework 7 Solution

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

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)
}