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Compiler Design

Description: Compiler Design
Number of Questions: 25
Created by: Aliensbrain Bot
Tags: Compiler Design Syntax Directed Translation IES (Indian Engineering Services) Syntax directed translation Parser Basics of Code Optimization Basics of code optimization Intermediate Code Generation Intermediate code generation Compiler Optimization

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Identify the odd one out.

Global common subexpression
Copy propagation
Register allocation
Code motion

Number of states for a grammar in SLR parser and LALR is

double of the other
somewhat dependent on each other
independent
dependent on grammar

Match the following. (Choose the most appropriate option) (A) S – Attributed Definition (i) Synthesized attribute (B) L ^_Attributed Definition (ii) Inherited attributed

A – i, B – ii
A – ii, B – i
A – i, B – i, B – ii
A – i, B – ii, A – ii

Consider the following grammar:A $\rightarrow$ BCC $\rightarrow$ +$A| \epsilon$B $\rightarrow$ id In the predictive parser table, m, of grammar the entire M [A, id] and M [C, $] respectively, is

{A $\rightarrow$ BC} and {C $\rightarrow$$\epsilon$}
{A $\rightarrow$ BC} and { }
{A $\rightarrow$ BC} and {C $\rightarrow$+A}
{B $\rightarrow$ id} and {C $\rightarrow$ $\epsilon$}

Expression corresponding to this DAG is

a + a * (b - c) + (b - c) *d
a + a × (b - c) + (b - c) * (e/f)
a*a + (b - c) + (b - c) * (e/f)
a + a* (b - c) + (b + c) * (e/f)

The grammar E $\rightarrow$E + E |E*E|(E)| id is

left recursive
ambiguous
both (1) and (2)
none of these

Which of the following statements is false about viable prefixes?

SLR passing is based on the fact that LR(0) automata recognize viable prefixes.
It is always possible to add terminal symbols to the end of a viable prefix to obtain a right- sentential form.
Both (1) and (2)
None of the above

Which of the following is generally not present in activation record?

Access link
List of all processes in system
Saved machine status
Control link

Match the following.


i. Token	a. Character i, f, letter followed by latter and digits etc
ii. Pattern	b. if, id, comparison
iii. Lexeme	c. if, pi, < =

i – b, ii – a, iii – c
i – b, ii – c, iii – a
i – c, ii – a, iii – b
i – c, ii – b, iii - a

A$\rightarrow$BCC $\rightarrow$ + B {print ('+'); }C|$\epsilon$ B$\rightarrow$ D* B {print ('*')i}|D D$\rightarrow$ (A)|id {print (id.value); } For an input '5 + 6*7' this translation scheme prints

5 + 6 *7
5 + *6 7
5 6 + 7*
5 6 7 *+

Consider the following statements. (i) Every SLR (1) grammar is unambiguous (ii) There are many unambiguous grammars that are not SLR (1)

i - True, ii - True
i - True, ii - False
i - False, ii - True
i - False, ii - False

Eliminate left recursion S ^{$\rightarrow$} Aa|b A ^{$\rightarrow$}Ac|Sd|f

$S \rightarrow Aa | b \\ A \rightarrow bdA'| fA' \\A | \rightarrow cA'| adA'| \epsilon$
$S \rightarrow Aa | b \\ A \rightarrow cdA'| dA' \\ A' \rightarrow bA'| adA' |\epsilon $
$S\rightarrow Aa|b \\ A\rightarrow bdA'|fA'$
$S \rightarrow Aa| b$ $A \rightarrow bdA'|fA'$ $A'\rightarrow cA'|adA'$

Computer value of 5 # 9 and 3 is

Find the false statement.

No left recursive or ambiguous grammar can be LL(1)
The class of grammars that can be parsed using LR methods is a proper subset of the class of grammars that can be the passed by LL method
LR parsing is non - back tracking method
LR methods can describe more languages than LL - grammars

Follow (E) is

{ )}
{$}
both (1) and (2)
none of the above

The grammar

is ambiguous
is unambiguous
cannot be decided
generates two parse trees for one input string

The grammar $S \rightarrow SA|A$ $A \rightarrow a$

SLR (1) but not LL (1)
SLR (1) and LL (1)
Not SLR (1), not LL (1)
Not SLR (1), but LL (1)

The value printed by the following is $ (( A + B ) * C + ( D * E) + ( F *G)) $

$ ++*+ABC*DE*FG $
$ +*++ABC*DE*FG $
$AB+C*DE*+FG*+$
$ABC+*DE*+FD*+$

While (i < = limit - 2) statement becomes $ \begin{cases} t=\text{limit - 2} \\ \text{While } (i \Leftarrow t) \end{cases} $ This optimization is caused by

copy propagation
code motion
dead code elimination
insufficient information

Consider the grammar. T$\rightarrow$T*F|F F $\rightarrow$ id For a sentence id₁ *id₂, the handlers in the right sequential form of the reduction are

id₁, F, id₂, T *F
id₁, id₂, T *F
id₁, F, T *F
id₁, F, id₂

Correct Option: A

Explanation:


Right sequential form	Handle	Reduction production
$id_{1}*d_2$	$id_1$	$F \rightarrow id$
$F * d_2$	F	$T \rightarrow F$
$T * d_2$	$id_2$	$F \rightarrow id$
$T * F$	$T * F$	$ E \rightarrow T * F $

Consider the grammar with transition rule and E is the start symbol.

E $\rightarrow$ E₁ $\ne$ T {E₁.value = E₁.value + T.value}
|T {E.value = T.value} T $\rightarrow$ T₁ & F {T.value = T₁.value / F.value}
|F {T.value = F.value} F $\rightarrow$ num {F.value = num.value} Concept E. The value for the root of the parse tree for expression 6 # 4 and 2 # 16 and 4 is

Consider the grammar.

First ( E' ) and first ( E ) are respectively

{+, ^$\epsilon$}, { , id }
{+ , ^$\epsilon$}, { ^*, ^$\epsilon$}
{^*, ^$\epsilon$}, {C, id}
none of these

Choose the most appropriate option. (i) Indirect triples is advantageous than triples. (ii) With indirect triple, an optimizing compiler can move an instruction by reordering the instruction list, without affecting the triples themselves.

i is true but ii is false.
i and ii are true and ii is the cause of i.
i and ii are true but ii is not the cause of i.
Both of them are false.

Left factor the grammar. $A \rightarrow ad|a|ab|abc|b$

$A \rightarrow aA'|b\\A' \rightarrow d| \epsilon |b|bc$
$A \rightarrow aA'| b\\A'\rightarrow d |\epsilon| bA\\A'\rightarrow \epsilon| c$
$A \rightarrow aA'| b\\A'\rightarrow d | bA'\\A\rightarrow c $
$A \rightarrow aA' | b\\A' \rightarrow d | b | bc$

Following are the production and the action.

$L\rightarrow E \eta$
$E \rightarrow \{ print('+'); \}E_1 + T$
$E \rightarrow R$
$E \rightarrow \{ print('+'); \}T_1 + F$
$T\rightarrow F$
$F\rightarrow (E)$
$F\rightarrow \text{digit \{print (digit lexval)\} }$
This is syntax directed translation for

prefix to infix
infix to prefix
postfix to infix
infix to postfix

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