Compilers
Question 1 |
Generation of intermediate code based on an abstract machine model is useful in compilers because
it makes implementation of lexical analysis and syntax analysis easier | |
syntax-directed translations can be written for intermediate code generation | |
it enhances the portability of the front end of the compiler | |
it is not possible to generate code for real machines directly from high level language programs |
Question 1 Explanation:
In Intermediate code optimizations can also enhances the probability of optimizer.
Question 2 |
Type checking is normally done during
lexical analysis | |
syntax analysis | |
syntax directed translation | |
code optimization |
Question 2 Explanation:
Type checking is normally done during syntax directed translation.
Question 3 |
The number of tokens in the Fortran statement DO 10 I = 1.25 is
3 | |
4 | |
5 | |
None of the above |
Question 3 Explanation:
DO → 1
10 → 2
I → 3
= → 4
1.25 → 5
10 → 2
I → 3
= → 4
1.25 → 5
Question 4 |
The number of tokens in the following C statement.
printf("i = %d, &i = %x", i, &i);
is
3 | |
26 | |
10 | |
21 |
Question 4 Explanation:
We have six different types of tokens are available
(i) Keyword
(ii) Identifier
(iii) Constant
(iv) Variable
(v) String
(vi) Operator
Print = Token 1
( = Token 2
"i=%d%x" = Token 3 [Anything inside " " is one Token]
, = Token 4
i = Token 5
, = Token 6
& = Token 7
i = Token 8
) = Token 9
; = Token 10
Here, totally 10 Tokens are present in the equation.
(i) Keyword
(ii) Identifier
(iii) Constant
(iv) Variable
(v) String
(vi) Operator
Print = Token 1
( = Token 2
"i=%d%x" = Token 3 [Anything inside " " is one Token]
, = Token 4
i = Token 5
, = Token 6
& = Token 7
i = Token 8
) = Token 9
; = Token 10
Here, totally 10 Tokens are present in the equation.
Question 5 |
For the program segment given below, which of the following are true?
program main (output);
type link = ^data;
data = record
d : real;
n : link
end;
var ptr : link;
begin
new (ptr);
ptr:=nil;
.ptr^.d:=5.2;
write ln(ptr)
end.
The program leads to compile time error | |
The program leads to run time error | |
The program outputs 5.2 | |
The program produces error relating to nil pointer dereferencing | |
None of the above |
Question 5 Explanation:
Note: Out of syllabus.
Question 6 |
Using longer identifiers in a program will necessarily lead to:
Somewhat slower compilation | |
A program that is easier to understand | |
An incorrect program | |
None of the above |
Question 6 Explanation:
Lexical analyzer will take more time to recognize the longer identifiers.
Question 7 |
In a compiler the module the checks every character of the source text is called:
The code generator. | |
The code optimizer. | |
The lexical analyser. | |
The syntax analyser. |
Question 7 Explanation:
Lexical analyzer phase checks every character of text to identify tokens.
Question 8 |
Consider a program P that consists of two source modules M1 and M2 contained in two different files. If M1 contains a reference to a function defined in M2, the reference will be resolved at
Edit time | |
Compile time | |
Link time | |
Load time |
Question 8 Explanation:
The link time can gives the reference to the executable file when the functions are present in the other modules.
Question 9 |
Match the following:
(P) Lexical analysis (1) Graph coloring (Q) Parsing (2) DFA minimization (R) Register allocation (3) Post-order traversal (S) Expression evaluation (4) Production tree
P-2, Q-3, R-1, S-4 | |
P-2, Q-1, R-4, S-3 | |
P-2, Q-4, R-1, S-3 | |
P-2, Q-3, R-4, S-1 |
Question 9 Explanation:
P) Lexical analysis is related with FA and Regular expressions.
Q) Expression can be evaluated with postfix traversals.
R) Register allocation can be done by graph colouring.
S) The parser constructs a production tree.
Hence, answer is ( C ).
Q) Expression can be evaluated with postfix traversals.
R) Register allocation can be done by graph colouring.
S) The parser constructs a production tree.
Hence, answer is ( C ).
Question 10 |
Match the following:
(P) Lexical analysis (i) Leftmost derivation (Q) Top down parsing (ii) Type checking (R) Semantic analysis (iii) Regular expressions (S) Runtime environments (iv) Activation records
P ↔ i, Q ↔ ii, R ↔ iv, S ↔ iii | |
P ↔ iii, Q ↔ i, R ↔ ii, S ↔ iv | |
P ↔ ii, Q ↔ iii, R ↔ i, S ↔ iv | |
P ↔ iv, Q ↔ i, R ↔ ii, S ↔ iii |
Question 10 Explanation:
Regular expressions are used in lexical analysis.
Top down parsing has left most derivation of any string.
Type checking is done in semantic analysis.
Activation records are loaded into memory at runtime.
Top down parsing has left most derivation of any string.
Type checking is done in semantic analysis.
Activation records are loaded into memory at runtime.
Question 11 |
Match the following according to input (from the left column) to the compiler phase (in the right column) that processes it:
P→(ii), Q→(iii), R→(iv), S→(i) | |
P→(ii), Q→(i), R→(iii), S→(iv) | |
P→(iii), Q→(iv), R→(i), S→(ii) | |
P→(i), Q→(iv), R→(ii), S→(iii) |
Question 11 Explanation:
Character stream is input to lexical analyzer which produces tokens as output. So Q → (iv).
Token stream is forwarded as input to Syntax analyzer which produces syntax tree as output. So, S → (ii).
Syntax tree is the input for the semantic analyzer, So P → (iii).
Intermediate representation is input for Code generator. So R → (i).
Token stream is forwarded as input to Syntax analyzer which produces syntax tree as output. So, S → (ii).
Syntax tree is the input for the semantic analyzer, So P → (iii).
Intermediate representation is input for Code generator. So R → (i).
Question 12 |
A lexical analyzer uses the following patterns to recognize three tokens T1, T2, and T3 over the alphabet {a,b,c}.
1: a?(b∣c)*aT
2: b?(a∣c)*bT
3: c?(b∣a)*c
Note that ‘x?’ means 0 or 1 occurrence of the symbol x. Note also that the analyzer outputs the token that matches the longest possible prefix.
If the string bbaacabc is processes by the analyzer, which one of the following is the sequence of tokens it outputs?
1: a?(b∣c)*aT
2: b?(a∣c)*bT
3: c?(b∣a)*c
Note that ‘x?’ means 0 or 1 occurrence of the symbol x. Note also that the analyzer outputs the token that matches the longest possible prefix.
If the string bbaacabc is processes by the analyzer, which one of the following is the sequence of tokens it outputs?
T1T2T3 | |
T1T1T3 | |
T2T1T3 | |
T3T3 |
Question 12 Explanation:
a? means either 0 or 1 occurrence of “a”, so we can write T1, T2 and T3 as:
T1 : (b+c)*a + a(b+c)*a
T2 : (a+c)*b + b(a+c)*b
T3 : (b+a)*c + c(b+a)*c
Now the string is: bbaacabc
Please NOTE:
Token that matches the longest possible prefix
We can observe that the longest possible prefix in string is : bbaac which can be generated by T3.
After prefix we left with “abc” which is again generated by T3 (as longest possible prefix).
So, answer is T3T3.
T1 : (b+c)*a + a(b+c)*a
T2 : (a+c)*b + b(a+c)*b
T3 : (b+a)*c + c(b+a)*c
Now the string is: bbaacabc
Please NOTE:
Token that matches the longest possible prefix
We can observe that the longest possible prefix in string is : bbaac which can be generated by T3.
After prefix we left with “abc” which is again generated by T3 (as longest possible prefix).
So, answer is T3T3.
There are 12 questions to complete.