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GATE | GATE-CS-2015 (Set 1) | Question 64
  • Difficulty Level : Easy
  • Last Updated : 20 Feb, 2015

The least number of temporary variables required to create a three-address code in static single assignment form for the expression q + r/3 + s – t * 5 + u * v/w is
(A) 4
(B) 8
(C) 7
(D) 9


Answer: (B)

Explanation:

The correct answer is 8. This question was asked as a fill in the blank type question in the exam.

 

Three address code is an intermediate code generated by compilers while optimizing the code. Each three address code instruction can have atmost three operands (constants and variables) combined with an assignment and a binary operator. The point to be noted in three address code is that the variables used on the left hand side (LHS) of the assignment cannot be repeated again in the LHS side. Static single assignment (SSA) is nothing but a refinement of the three address code.

So, in this question, we have

t1 = r / 3;

t2 = t * 5;

t3 = u * v;

t4 = t3 / w;

t5 = q + t1;

t6 = t5 + s;

t7 = t6 - t2; 

t8 = t7 + t4;

Therefore, we require 8 temporary variables (t1 to t8) to create the three address code in static single assignment form.

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