compiler design lab viva questions and answers

I want compiler viva questions with answers ?
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1) For the C program given below the statement that does not hold true is

for (i = 0; ifor(j = 0; jif (i%c)
{
x += (4*j+5*i);
y+=(7+4*j);
}
}
}

A) There is a scope strength reduction
B) There is a scope of dead code elimination
C) There is a scope of common sub-expression elimination
D) None of the above

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(2) In compilers generation of intermediate code based on an abstract machine model is useful because

(A) Syntax-directed translations can be written for intermediate code generation
(B) To generate code for real machines directly from high-level language programs is not possible
© Portability of the front end of the compiler is enhanced
(D) Implementation of lexical and syntax analysis is easier

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(3) Consider the grammar where P, Q, R are not terminals and r, s, t are terminals

a. P->Q R
b. P->Q s R
c. P->ε
d. P->Q t R r

The grammar rules that violate the requirements of an operator grammar is

(A) a and c only
(B) b and c only
© a and d only
(D) a only

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(4) Which one of the following statement is false for the SLR (1) and LALR (1) parsing tables for a context free grammar?

(A) The reduce entries in both the tables may be different
(B) The error entries in both the tables may be different
© The go to part of both tables may be different
(D) The shift entries in both the tables may be identical

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(5) We have a grammar with not epsilon and unit production (i.e. of type A->ε and
A ->a) to parse a string, with n tokens. What is the maximum number of reduces moves that can be taken by a bottom-up parser for this grammar?

(A) 2n - 1
(B) 2n
© n-1
(D) n/2

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For questions 6 and 7 refer to the data given below:

The processor allows only register operands in its instructions and the given code segment is executed in that processor. For each instruction almost two source operands and one destination

operand is available. Assume that all the variables are dead after this code segment.

z = x + y;
p = z * x;
q = c + x;
j = z * z;

If (j > x)
{
M = x*x;
}
else
{
p = p*p;
q = q*q;
}

(6) Only two registers are available in the instruction set architecture of the processor. The code motion moves the statements from one place to another while preserving correctness. The only allowed complier optimization is code motion. In the compiled code, the minimum number of spills to memory is

(A) 0
(B) 1
© 2
(D) 3

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(7) Assume that no other optimization other than optimizing register allocation is applied. To compile this code segment without any spill to memory the minimum number of registers needed in the instruction set architecture of the processor is

(A) 3
(B) 6
© 4
(D) 5

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(8) We have the grammar E->E + n I E x n I n. The handles in the right-sentential form of the reduction for a sentence n + n x n are

(A) n, n + n and n + n x n
(B) n, E + n and E x n
© n, E + n and E + E x n
(D) n, E + n and E + n x n

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(9) The languages that need heap allocation in the runtime environment are

(A) Those that use global variables
(B) Those that use dynamic scoping
© Those that support recursion
(D) Those that allow dynamic data structure

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(10) When is the type checking usually done?

(A) During syntax directed translation
(B) During lexical analysis
© During code optimization
(D) During syntax analysis

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(11) What information need to be included in an object module, if a linker is given object modules for a set of programs that were compiled separately?

(A) Names and locations of all external symbols defined in the object module
(B) Object code
© Absolute addresses of internal symbols
(D) Relocation bits

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(12) We have two sets of LR (1) items of LR (1) grammar described below.

X->c.X, c/d X->c.X, $
X->.cX, c/d X->.cX, $
X->.d, c/d X->.d, $

Related to the merging of two sets in the corresponding parser, the statement that does not hold true is

(A) Cannot be merged since look aheads are different.
(B) Cannot be merged but will result in R-R conflict
© Cannot be merged but will result in S-R conflict
(D) All of the above

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(13)
a. Begin
b. Program
c. <>

While compiling a Pascal program, without looking at the next input character the string that can be definitely said to be token is

(A) b
(B) c
© a
(D) All of the above

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(14)
S->xx W {print”1”}
S->y {print”2”}
W->Sz {print”3”}

A shift reduce parser carries out the actions specified within braces immediately after reducing with the corresponding rule of grammar. Using the syntax directed translation scheme described

by the above rule, the translation of xxxxyzz is

(A) 11231
(B) 11233
© 23131
(D) 33211

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(15) In some programming language, L denotes the set of letters and D denotes the set of digits. An identifier is permitted to be a letter followed by any number of letters or digits. The expression that defines an identifier is

(A) (L.D)*
(B) (L + D)*
© L (L.D)
(D) L (L + D)*

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(16) Which one of the following statement is true?

(A) Canonical LR parser is more powerful than LALR parser
(B) SLR parser is more powerful than LALR
© LALR parser is more powerful than canonical LR parser
(D) SLR parser, canonical LR parser and LALR parser all have the same power

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(17) The activities are listed below. What is the pass numbers of each of the activities respectively?

a. Object code generation
b. Literals added to the literal table
c. Listing printed
d. Address resolution of local symbols that occur in a two pass assembler

(A) 1, 2, 2, 2
(B) 2, 1, 1, 1
© 1, 2, 1, 2
(D) 2, 1, 2, 1

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(18) In a compiler _______________ checks every character of the source text.

(A) The lexical analyzer
(B) The syntax analyzer
© The code generator
(D) The code optimizer

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For questions 19 and 20 refer to the data given below:

The programming language given below is written in the programming language that does not allow nested declarations of functions and allows global variables.
global int j = 100, k = 5;
void M(n)
{
int j = 10;
print (n + 10);
j = 200;
k = 20;
print (n);
}
main()
{
M(j + k);
}

(19) What is the output of the above program if the programming language uses static scoping and call by need parameter passing mechanism?

(A) 25, 220
(B) 115, 105
© 220, 115
(D) 105, 220

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(20) What is the output of the above program if the programming language uses dynamic scoping and call by name parameter passing mechanism?

(A) 105, 25
(B) 105, 115
© 220, 25
(D) 115, 220

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(21) ______________ is not an advantage of using shared, dynamically linked libraries as opposed to using statically linked libraries.

(A) Lesser overall page fault rate in the system
(B) Faster program startup
© Existing programs need not be relinked to take advantage of newer versions
(D) Smaller sizes of executable files

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(22) The instructions of a simplified computer, which has only two registers, are given below:
OP Rj, Rk - Performs Rj OP Rk and stores the result in register Rk.
OP m, Rj - Performs the content of memory location m OP Rj and stores the result in Rj
MOV m, Rk - Moves the content of memory location m to register Rk.
MOV Rk, m - Moves the content of register Rk to memory location m.
OP is either ADD or SUB.

We have the following basic block:

T1 = a + b
T2 = c + d
T3 = e - T2
T4 = T1 - T3

Assuming that all operands are initially in memory and the final value of the computation in memory, the minimum number of MOV instructions in the code generated for this basic block is

(A) 3
(B) 2
© 6
(D) 4

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For questions 23 and 24 refer to the data given below:

Consider the language L = {ai bi I i ? j}?

(23) The grammar that generates this language is

(A) S->aS\Sb\a\b
(B) S->AC\CB
C->aC b\a\b
A->aA\ ∈
B->Bb\ ∈
© S->AC\CB
C->aCb\∈
A->aA\a
B->bB\b
(D) S->AC\CB
C->aC b\∈
A->aA\ ∈
B->Bb\ ∈

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(24) In the correct grammar from question 23 to generate the string al bm with l ≠ m, the length of the derivation (number of steps starting from s) is

(A) max (l, m) + 3
(B) max (l, m) + 2
© l + m + 2
(D) l + m + 3

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(25) Consider the following C program:

int min (){ /*line 1*/
int I, N; /*line 2*/
fro (I=0, I
While creating the object module, the compiler’s response about line no. 3 is

(A) Only syntax error
(B) No compilation error
© Only lexical error
(D) Both lexical and syntax error

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(26) We have the translation scheme given below:

S->FR
R->*E{print(‘*’);R\ε
E->F + E{print (‘+’);\F
F->(S)\id {print (id.value);}

In the above translation scheme id represents the token in integer form and id value represents the corresponding integer value. What will be printed by this translation scheme when an input

is ‘2 * 3 + 4’?

(A) 2 3 * 4 +
(B) 2 3 4 + *
© 2 * + 3 4
(D) 2 * 3 + 4

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(27) We have the grammar with the translation rules:

E->E1 # T {E.value = E1.value * T.value}
I T {E.value = T.value}
T->T1 & F {T.value = T1.value F.value}
I F {T.value = F.value}
F->num {F.value = num.value}

In the above grammar E is the start symbol. What will be the value of E for the root parser tree for the expression 2 # 3 & 5 # 6 & 4?

(A) 50
(B) 100
© 200
(D) 160

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(28) For the expression grammar

E->E*F I F+E I F
F->F – I id

The statement, which holds true, is

(A) + and – have same precedence
(B) Precedence of * is higher +
© Precedence of - is higher *
(D) Precedence of + is higher *

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(29) Which one of the following statement holds true for a bottom-up evaluation of syntax directed definition?

(A) Inherited attributes can always be evaluated
(B) Inherited attributes can never be evaluated
© Inherited attributes can be evaluated only if the definition is L-attributed
(D) Inherited attributes can be evaluated only if the definition has synthesized attributes

For questions 30 and 31 refer to the data given below:

{S, A, B} is the non-terminal alphabet and {a, b} is the terminal alphabet of the CFG. S is the start symbol. The set of production rules are given below,
S->aB S->bA
B->b A->a
B->bS A->aS
B->aBB A->bAA

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(30) The string that is generated by the grammar is

(A) aabbbb
(B) abbbba
© aabbab
(D) aaaabb .