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GATE | GATE CS 2011 | Question 36
  • Last Updated : 19 Nov, 2018

Consider evaluating the following expression tree on a machine with load-store architecture in which memory can be accessed only through load and store instructions. The variables a, b, c, d and e initially stored in memory. The binary operators used in this expression tree can be evaluate by the machine only when the operands are in registers. The instructions produce results only in a register. If no intermediate results can be stored in memory, what is the minimum number of registers needed to evaluate this expression?

 

gate2011Q26
(A) 2
(B) 9
(C) 5
(D) 3


Answer: (D)

Explanation:

R1←c,  R2←d,  R2←R1+R2,  R1←e,  R2←R1-R2
Now to calculate the rest of the expression we must load a and b into the registers but we need the
content of R2 later.
So we must use another Register.
R1←a, R3←b, R1←R1-R3, R1←R1+R2

Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html


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