Compiler Design Test 1

Result

Compiler Design Test 1

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If C is a programming language, then which of the following cannot be a token produced by lexical analysis phase?

A
++

B
*=

C
<<

D
None of these
Solution
- The first phase of compiler is called lexical analysis or scanning. The lexical analyzer reads the stream of characters making up the source program and groups the characters into meaningful sequence called lexemes. For each lexeme, the lexical analyzer produces as output a token of the form that it passes on to the subsequent phase, syntax analysis.
- Tokens may be of the form of identifiers, keywords, constants etc. from the input program.
++
unary operator (incremental operator)
*=
Assignment operator
<<
Bitwise operator

++, *=, << are lexemes that is mapped into the token < ++ >, < *= > and < << >
All operators are token
Therefore option 4 is correct.
Question 2
1 / -0

If below given options are written in C language, then which of the following code can produce lexical error?

A
register float x = int;

B
char y = y;

C
short int a = 3.12;

D
int 2y;
Solution
- The first phase of compiler is called lexical analysis or scanning. The lexical analyzer reads the stream of characters making up the source program and groups the characters into meaningful sequence called lexemes. For each lexeme, the lexical analyzer produces as output a token of the form that it passes on to the subsequent phase, syntax analysis.
- Tokens may be of the form of identifiers, keywords, constants etc. from the input program.
- In C language identifiers are the names given to variables, constants, functions and user-define data. Identifier cannot start with number.
- Since 2y is not valid identifier. Therefore, lexical analysis will produce an error
Important Points:
All 4 statement will produce error, but lexical error can be produced only with option 4.
Question 3
1 / -0

Which of the following statements is/are TRUE about the lexical analyzer?
I. It reads the input characters and produces a sequence of tokens that the parser uses for syntax analysis.
II. Lexical analyzer removes comments and white spaces in the form of blank, tab, and newline characters.
III. It correlates error messages from the compiler with the source program.

A
I

B
I and II

C
I, II and III

D
I and III
Solution
- Lexical analyzer is the first phase of a compiler.
- The input is a high-level language program, such as a ’C’ program in the form of a sequence of characters.
- The output is a sequence of tokens that are sent to the parser for syntax analysis Strips off blanks, tabs, newlines, and comments from the source program Keeps track of line numbers and associates error messages from various parts of a compiler with line numbers.

Question 4

1 / -0

The number of tokens in the following C program

int main() {
float p = 2.33;
double q = 19;
if(p != q)
printf("p = %f", p++);
else
printf("q = %lf", ++q);

}

What is the number of tokens?

39-39

394-391

397-396

399-392

Solution

Line no.	Tokens	Number of Tokens
Line 1	int \| main \| ( \| ) \| } \|	5
Line 2	float \| p \| = \| 2.33 \| ; \|	5
Line 3	double \| q \| = \| 19 \| ; \|	5
Line 4	printf \| ( \| "p = %f" \| , \| p \|++ \| ; \|	8
Line 5	if \| ( \| p \| != \| q \| ) \|	6
Line 6	else \|	1
Line 7	printf \| ( \| "q = %lf" \| , \| ++ \| q \| ; \|	8
Line 8	} \|	1

Total Number of tokens = 5 + 5 + 5 + 8 + 6 + 1 + 8 + 1 = 39

Notes:

| is used to separate the token

Question 5
1 / -0

Consider the below given table in which regular expression is mapped to unique token:
Regular Expression Token
p*(p | q)p* 3
qp* 1
p+(r | q) pq 2
What will be the output when the string ''pppqppqprpqp" is scanned by the lexical analyzer if the analyzer outputs the token that matches the longest possible prefix if it is mandatory to use all the tokens at least once?

A
3122

B
3213

C
2133

D
3123

Solution

String: pppqppqprpqp
Regular expression
p*(p | q)p*
qp*

p+(r | q) pq

p*(p | q)p*
Token
3
1
2
3
String
pppqpp
q
prpq
p

The output is 3123
Question 6
1 / -0

Consider the following statements:
S1: New entries are created in symbol table for each new identifier during lexical analysis.
S2: Lexical analyzer detects ill-formed numeric literals and input characters that are not in the source language.
Which of the following is true?

A
Only S1

B
Only S2

C
Both S1 and S2

D
Neither S1 nor S2

Solution

The correct answer is option 3 i.e. Both S1 and S2.
A lexical analyzer creates a symbol-table entry as soon as it sees the characters that make up a lexeme. Also, it detects ill-formed numeric literals and input characters that are not in the source language.
Question 7
1 / -0

Consider the below-given code written in C language
#include<stdio.h>
int main()
{
int k = 20;
whlie(--k)
printf(“GATE CS 2021\n”);
}
Identity the compiler’s response about the above-given code while creating the object- module

A
Lexical Error

B
Prints GATE CS 2021 19 times

C
Prints GATE CS 2021 20 times

D
Syntactic Error

Solution

#include<stdio.h>
int main()
{
int k = 20;
whlie(--k) \\incorrect while spelling
printf(“GATE CS 2021\n”);
}
while is a keyword in C programming language but code has incorrect spelling of while, that is, whlie
Now, it will be an error
Lexical analyzer will treat whlie as a valid identifier
But the syntax analyzer will catch this error. Hence it is a syntactic error.
Question 8
1 / -0

A particular BNF definition for a “word” is given by the following rules.
<word> :: = <letter> | <letter><charpair> | <letter><intpair>
<charpair> :: = <letter><letter> | <charpair><letter><letter>
<intpair> :: = <integer><integer> | <intpair><integer><integer>
<letter> :: = A | B | C | ….. | Y | Z
<integer> :: = 0 | 1 | 2 | …. | 9
Which of the following lexical entries can be derived from < word >?
I. TESTBOOK
II. TTP
III. T20

A
I, II and III

B
Only I and II

C
Only II and III

D
Only II

Solution

<word> :: = <letter><charpair>
<word> :: = <letter><charpair><letter><letter>
<word> :: = <letter><charpair><letter><letter><letter><letter>
<word> :: = <letter><charpair><letter><letter>l<etter><letter>
<word> :: = <letter><charpair><letter><letter>letter><letter>l<etter><letter> // cannot derive TESTBOOK
Since number of letter in TESTBOOK = 8
<word> :: = <letter><letter><letter><letter><letter>letter><letter>letter><letter> // cannot derive TESTBOOK
<word> :: = <letter><charpair>
<word> :: = <letter> <etter><letter>
<word> :: = TTP //derived
<word> :: = <letter><intpair>
<word> :: = <letter><intpair>
<word> :: = <letter><integer><integer>.
<word> :: = T20 //derived
Option 3 is correct

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

++	unary operator (incremental operator)
*=	Assignment operator
<<	Bitwise operator

Regular Expression	Token
p(p \| q)p	3
qp*	1
p+(r \| q) pq	2

Compiler Design Test 1

Result

Compiler Design Test 1

Score

-

Rank

-

SHARING IS CARING

Question 1

++

*=

<<

None of these

Solution

Question 2

register float x = int;

char y = y;

short int a = 3.12;

int 2y;

Solution

Question 3

I

I and II

I, II and III

I and III

Solution

Question 4

39-39

394-391

397-396

399-392

Solution

Question 5

3122

3213

2133

3123

Solution

Question 6

Only S1

Only S2

Both S1 and S2

Neither S1 nor S2

Solution

Question 7

Lexical Error

Prints GATE CS 2021 19 times

Prints GATE CS 2021 20 times

Syntactic Error

Solution

Question 8

I, II and III

Only I and II

Only II and III

Only II

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.