Compiler Design Test 4

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Question 1/10
1 / -0

Generation of intermediate code based on an abstract machine model is useful in compilers because:
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A
Syntax translations are easier for intermediate code generation

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B
It is difficult to generate executable code from high level language program.

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C
It enhances the portability of the compiler system program

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D
It makes implementation of lexical and syntax analysis easier
Solutions

Explanation:
Intermediate code is:
- Machine independent code so a compiler can optimize it without worrying about the architecture on which the code is going to execute.
- Main purpose of intermediate code generation is to enhance the portability of the compilers,
hence option 3 is the most suitable option here.
Question 2/10
1 / -0

Consider the following the statements
S1: Static allocation bindings do not change at runtime
S2: Heap allocation allocates and de-allocates storage at run time
Which of the following statements is/are true?
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A
S1 is true and S2 is false

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B
S2 is true and S1 is false

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C
Both S1 and S2 are true

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D
Both S1 and S2 are false
Solutions
- Dynamic binding changes at run time not static.
- Heap allocation supports dynamic data structures because it allocates and de-allocates storage at run time.
Question 3/10
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Consider the productions for the below given grammar?
S → Q | S # Q
Q → Q # P | P
P → R @ P | R
R → id
Which of the following is false?

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A
S is left recursive and # has lowest order of precedence

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B
P is right recursive and @ has highest order of precedence

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C
P is right recursive and @ is right associative

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D
Q is right recursive and # is left associative

Solutions

From grammar, it can be said that
Associativity
Since S is left recursive therefore # is left-associative
Since Q is left recursive therefore # is left-associative
Since P is right recursive therefore @ is right-associative
Order of preference:
# has the lowest order of precedence
@ has the highest order of precedence
Question 4/10
1 / -0

Consider the following C code fragment:
void main()
{
…
x()
}
int x()
{
…
y()
}
int y()
{
…
}
Which of the following statement is correct about activation record?

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A
Before executing main activation record of x is created.

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B
Before executing y, activation record of main and x is pushed in stack.

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C
Before executing y, activation record of main and x is popped from stack.

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D
Before executing x, activation record of main and y is pushed in stack.

Solutions
Procedure calls and returns are usually managed by a run-time stack called the control stack. Each live activation has an activation record (sometimes called a frame) on the control stack, with the root of the activation tree at the bottom, and the entire sequence of activation records on the stack corresponding to the path in the activation tree to the activation where control currently resides. The latter activation has its record at the top of the stack.
Question 5/10
1 / -0

Consider the following statements:
(i) The symbol table contains information about the lexical and the syntactical structure of the program.
(ii) Output of the semantic analysis is the annotated parse tree.
(iii) The intermediate code generator does not use the symbol table.
(iv) The intermediate code generation and the machine code generation can never be combined in a single step.
Which of the above statement/s is/are TRUE about the compilation?

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A
(i) and (ii) only

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B
(ii) and (iii) only

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C
(i) and (iv) only

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D
All are true

Solutions

(i) True: The symbol table contains information about the lexical and the syntactical structure of the program.
(ii) True: Output of the semantic analysis is the annotated parse tree.
(iii) False: The intermediate code generator does use the symbol table.
(iv) False: The intermediate code generation and the machine code generation can be combined in a single step.
Question 6/10
1 / -0

Consider the following code segment:
p = q – r
s = r + t
p = s + p
s = a + s
What it the minimum number of total variables required to convert the above code segment to static single assignment?

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A
8-8

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B
84-81

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C
87-86

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D
89-82

Solutions

In compiler design, static single assignment form (SSA) is a property of an intermediate representation, which requires that each variable is assigned exactly once, and every variable is defined before it is used.
t1 = q – r
t2 = r + t
t3 = t2 + t1
t4 = a + t2
Variables are: t1, q, r, t2, t, t3, t4, a
Therefore, 8 variables are present in static single assignment form.
Question 7/10
1 / -0

Consider the statement
do
{
i = i + 1;
}
While(a[i] < b);
The minimum number of temporary variables required in three address code of the above statement is _____.

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A
2

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B
3

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C
4

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D
5

Solutions

t1 = i + 1
i = t1
t2 = i * 8 // refers memory allocation for given data type
t3 = a[t2]
if t3 < b
Hence, 3 variables are required in three address code.
Question 8/10
1 / -0

Choose the equivalent prefix form of the following expression
(a + (b − c))* ((d − e)/(f + g − h))

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A
* +a − bc /− de − +fgh

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B
* +a −bc − /de − +fgh

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C
* +a − bc /− ed + −fgh

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D
* +ab − c /− ed + −fgh

Solutions

Using given prefix form we can draw the expression tree as follows:
* is the root.
+ is its left child and / is right.
a is the left child of + and - is the right.
b and c are the left and right child of - respectively.
/ has - at both left and right child node.
left child node of / has d and c as left and right child respectively.
right child node of / has + and h as the left and right child respectively.
+ has f and g as left and right child respectively.
Using this tree we can deduct that prefix form of this tree is * +a − bc /− de − +fgh.
Question 9/10
1 / -0

Consider the grammar defined by the following production
P → Q | P * Q
Q → R + Q | R
R → T – R | T
T → id
where * is multiplication, + is addition and – is subtraction operation.
The output of the given expression 9 * 5 * 11 – 10 – 8 + 3 is _____.

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A
540-540

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B
5404-5401

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C
5407-5406

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D
5409-5402

Solutions

From the grammar it can be said that
Since P is left recursive therefore * is left associative
Since Q is right recursive therefore + is right associative
Since R is right recursive therefore – is right associative
Also, - has highest order of precedence and * has lowest order of precedence
Therefore
(9 * 5) * ((11 – (10 – 8)) + 3)
= 45 × 12 = 540
Question 10/10
1 / -0

Consider the following table:
A.
Activation record
p.
Linking loader
B.
Location counter
q.
Garbage collection
C.
Reference counts
r.
Subroutine call
D.
Address relocation
s.
Assembler
Matching A, B, C, D in the same order gives:

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A
p, q, r, s

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B
q, r, s, p

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C
r, s, q, p

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D
r, s, p, q

Solutions

Activation record - Subroutine call
Location counter- Assembler
Reference counts- Garbage collection
Address relocation- Linking loader

A.	Activation record	p.	Linking loader
B.	Location counter	q.	Garbage collection
C.	Reference counts	r.	Subroutine call
D.	Address relocation	s.	Assembler

Question 1/10

Syntax translations are easier for intermediate code generation

It is difficult to generate executable code from high level language program.

It enhances the portability of the compiler system program

It makes implementation of lexical and syntax analysis easier

Question 2/10

S1 is true and S2 is false

S2 is true and S1 is false

Both S1 and S2 are true

Both S1 and S2 are false

Question 3/10

S is left recursive and # has lowest order of precedence

P is right recursive and @ has highest order of precedence

P is right recursive and @ is right associative

Q is right recursive and # is left associative

Question 4/10

Before executing main activation record of x is created.

Before executing y, activation record of main and x is pushed in stack.

Before executing y, activation record of main and x is popped from stack.

Before executing x, activation record of main and y is pushed in stack.

Question 5/10

(i) and (ii) only

(ii) and (iii) only

(i) and (iv) only

All are true

Question 6/10

8-8

84-81

87-86

89-82

Question 7/10

2

3

4

5

Question 8/10

* +a − bc /− de − +fgh

* +a −bc − /de − +fgh

* +a − bc /− ed + −fgh

* +ab − c /− ed + −fgh

Question 9/10

540-540

5404-5401

5407-5406

5409-5402

Question 10/10

p, q, r, s

q, r, s, p

r, s, q, p

r, s, p, q