Computer Organization and Architecture -CODE SEGMENT

The following code segment is executed on a processor which allows only register operands in its instructions. Each instruction can have atmost two source operands and one destination operand. Assume that all variables are dead after this code segment.

   c = a + b;
   d = c * a;
   e = c + a;
   x = c * c;
   if (x > a) {
      y = a * a;
   }
   else {
     d = d * d;
     e = e * e;
  }

Suppose the instruction set architecture of the processor has only two registers. The only allowed compiler optimization is code motion, which moves statements from one place to another while preserving correctness. What is the minimum number of spills to memory in the compiled code?

(A) 0
(B) 1
(C) 2
(D) 3


Answer: (B)

Explanation:

r1......r2
a.......b......c = a + b
a.......c......x = c * c
a.......x......but we will have to store c in mem as we don't know if x > a
................. or not
y.......x......y = a * a
choosing the best case of x > a , min spills = 1