Compiler Design #2

Question

Consider translation rules :   S → S1 * A {S. Val = S1. Val + A. Val} /A {S. Val = A. Val} A → A1 – B {A. Val = A1. Val – B. Val} /B {A. Val = B. Val} B → C | B1 {B. Val = C. Val / B1. Val} /C {B. Val = C. Val} C → id {C. Val = Id. Val} Let say output for input 8/4 * 2 – 6 * 4 is a1 and output for input 6 * 2 / 1 – 2 * 3 is a2, then a1 – a2 is ___________.

Comments

a1 = 2

a2 = 9

so a1-a2 = 2-9 = -7

Answer 1

In SDT above the operator that occurs at lowest level has highest precedence and one that occurs nearest to start symbol has lowest precedence.

Precedence order of operators is $/ \gt - \gt *$. But $*$ operation here means addition. Thus all divisions take place first, then subtraction and then Addition.

So, $a_1 = 8/4 * 2 – 6 * 4$ evaluates as $(8/4)+(2-6)+4 = 2 + (-4) + 4 = 2$

And $a_2 = 6 * ((2 / 1) – 2) * 3  = 2 + (2-2) + 3 = 5$

$a_1 -a_2 = 2 - 5 = -3$

Comments

@mcjoshi. You did it wrongly. When its * operator, operation done is +. See the translation.

So, a1 = 2 and a2=5

Hence, a1 - a2 = -3

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Thanks for correcting me:)
Went according to answer instead of properly reading.

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6∗((2/1)–2)∗3=2+(2−2)+3=5a2=6∗((2/1)–2)∗3=2+(2−2)+3=5

Shouldn't it be 6+(2-2)+3=9

So the overall answer should be 2-9=-7