Question

Consider the following ANSI C code segment:

$z=x+3+y\to f1+y\to f2;$

$for(i=0;i<200;i=i+2)\{$

$if(z>i)\{$

$p=p+x+3;$

$q=q+y\to f1;$

$\left\}else\right\{$

$p=p+y\to f2;$

$q=q+x+3;$
}
}
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 $\to$ f1 or y $\to$ f2) in the optimized code, respectively, are :

• A. 403 and 102
• B. 203 and 2
• C. 303 and 102
• D. 303 and 2

Answer 1

The correct option is D 303 and 2

$t_1=y\to f_1\left(1dereference\right)$

$t_2=y\to f_2\left(1dereference\right)$

$t_3=x+3\left(1add\right)$

$z=t_3+t_1+t_2\left(2additions\right)$

For (i = 0; i < 200; i + = 2)
{
if (z > i)
{
p = p + $t_3$
q = q + $t_1(2add$)
}
else
{
p = p + $p_2$
q = q + $t_3$ (2 add)
}
}
If else condition $\Rightarrow$
Either if is executed (or) else is executed.
$\Rightarrow$ At any iteration 2 addition operations will be executed.
So, in loop the iterations are $\left(\frac{200}{2}\right)=100times.$
$\therefore$ In loop the number of additions = 100 × 2 = 200 additions
$\therefore$ Total additions = 200 + 1 + 2 + 100 loop additions (inside for loop) = 303
and 2 dereferences.
$\therefore$ Correct answer is 303 and 2 which is option (a).