1
GATE CSE 2003
MCQ (Single Correct Answer)
+1
-0.3
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?
A
Removing left recursion alone
B
Factoring the grammar alone
C
Removing left recursion and factoring the grammar
D
None of the above
2
GATE CSE 2003
MCQ (Single Correct Answer)
+2
-0.6

Consider the translation scheme shown below

$$\eqalign{ & S \to TR \cr & R \to + T\left\{ {pr{\mathop{\rm int}} (' + ');} \right\}\,R\,|\,\varepsilon \cr & T \to num\,\left\{ {pr{\mathop{\rm int}} (num.val);} \right\} \cr} $$

Here num is a token that represents an integer and num.val represents the corresponding integer value. For an input string '9 + 5 + 2', this translation scheme will print

A
9 + 5 + 2
B
9 5 + 2 +
C
9 5 2 + +
D
+ + 9 5 2
3
GATE CSE 2003
MCQ (Single Correct Answer)
+2
-0.6
Which of the following is NOT an advantage of using shared, dynamically linked libraries as opposed to using statically linked libraries?
A
Smaller sizes of executable files
B
Lesser overall page fault rate in the system
C
Faster program startup
D
Existing programs need not be re-linked to take advantage of newer versions of libraries
4
GATE CSE 2003
MCQ (Single Correct Answer)
+2
-0.6

Consider the syntax directed definition shown below.

GATE CSE 2003 Compiler Design - Code Generation and Optimization Question 15 English

Here, gen is a function that generates the output code, and newtemp is a function that returns the name of a new temporary variable on every call. Assume that ti's are the temporary variable names generated by newtemp. For the statement 'X : = Y + Z', the 3-address code sequence generated by this definition is

A
$$X = Y + Z$$
B
$${t_1} = Y + Z;X = {t_1}$$
C
$${t_1} = Y;{t_2} = {t_1} + Z;X = {t_2}$$
D
$${t_1} = Y;{t_2} = Z;\,{t_3} = {t_1} + {t_2};X = {t_3}$$
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