Consider the following ANSI C code segment:z = x + 3 + y->f1 + y->f2; for (i = 0; i < 200; i = i + 2){ if (z > i) { p = p + x + 3; q = q + y->f1; } else { p = p + y->f2; q = q + x + 3; } }Assume that the variable y points to a struct (allocated on the heap) containing two fields f1 and f2, and the local variables x, y, z, p, q, and i are allotted registers. Common sub-expression elimination (CSE) optimization is applied on the code. The number of addition and dereference operations (of the form y->f1 or y->f2 ) in the optimized code, respectively, are:

### 1. Analysis of Expressions Before and Inside the Loop

Before the loop:

• The sub-expressions evaluated are x + 3, y->f1, and y->f2.

• z = ((x + 3) + y->f1) + y->f2 requires $3$ additions and $2$ dereferences (y->f1 and y->f2).

Using Common Sub-expression Elimination (CSE) and Loop Invariant Code Motion:

• The expressions x + 3, y->f1, and y->f2 do not change throughout the execution of the loop because x, y, y->f1, and y->f2 are never modified.

• Thus, the compiler stores the results of x + 3, y->f1, and y->f2 in temporary registers before the loop.

### 2. Operations Inside the Loop

• The loop executes for values $i = 0, 2, 4, \dots, 198$, giving a total of $\frac{200 - 0}{2} = 100$ iterations.

• In each iteration:

  • The if branch or else branch is executed. Both branches perform exactly $2$ additions (e.g., p = p + (x + 3) and q = q + (y->f1) using precomputed values).

  • The loop increment i = i + 2 performs $1$ addition.

  • Hence, each iteration requires $2 + 1 = 3$ additions.

  • Across $100$ iterations: $100 \times 3 = 300$ additions.

  • No further pointer dereferences are executed inside the loop ($0$ dereferences).

### 3. Total Operations

• Total Additions:$3 \text{ (outside)} + 300 \text{ (inside)} = 303$

• Total Dereferences:$2 \text{ (outside)} + 0 \text{ (inside)} = 2$

Therefore, the optimized code has 303 additions and 2 dereferences, which corresponds to option (D).