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CS606  Current MID Term Papers Fall 2012 Date: 08-December-2012 to 19-December-2012

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Noboy have Current midterm paper of CS606 ................?????
Reply plxx.....

McQS 15 yahn jo pa jo mcqs han in main se hi thy

Q#1: how linker playes an important role with respect to compiler construction.(2 marks)

Q#2:define 2 pass compiler (2  marks)consider

Q#3: Draw LL(1) parsing table for the following grammar (5 marks)

S (arrow)aSlAb

A(arrow)XXZlE

X(arrow)dsl€

Y(arrow)ds€

Z(arrow)es

Q#4consider the following grammar (5 marks)

Statement(arrow)if expression the statement.

Statement (arrow)if expression the statement.

You are required provide an alternate production so that it may become from the backtracking.

Q#5:consider LL parse table(3 marks)

 

a

b

$

S

S(arrow)aB

S(arrow)abB

S(arrow)ab

B

 

B(arrow)b

B(arrow)€

Q#6: consider the following  grammar for arithmetic expressionsJ (3marks)

E(arrow)idQ

Q(arrow)ERl€

R(arrow)+Ql-l*Ql’Q

 syeda Areeba Thanks for sharing ur paper 

Note for All Members: You don’t need to go any other site for current Mid Term papers fall 2012, Because All discussed data/sharing of our members in this discussion are going from here to other sites. You can judge this at other sites yourself. So don’t waste your precious time with different links.

CS606

16-12-2012

Question No 1

What is the role of Automatic Code Generator with respect to compiler?

 

Question No2

How Linker play an important role with respects to compilers constructions?

 

Question No3

Consider the grammar for arithmetic expressions?

E→ id Q

Q→ ER│ ε

R→ +Q │-Q│*Q│/Q

Show the follow set for all the non-terminal in the grammar

 

Question No 4

Consider the following grammar. Calculate the first set of non- terminal S, A and B

S→ AB

A→ a│ε

B→ b│ε

 

Question No 5

Consider the following grammar. You are required to wrote leftmost derivation of string “aebb”

S→aS│Ab

A→XYZ│ε

X→cS│ε

Y→dS│ε

Z→eS

 

Question No 6

Considering string “ab” shows that the following grammar is ambiguous using parse tree. The ε below is epsilon

S→ABAB

A→aA│ε

B→bB│ε

 

 

By Oheena Shah

Attachments:

Oheena Shah Thanks for sharing 

Note for All Members: You don’t need to go any other site for current Mid Term papers fall 2012, Because All discussed data/sharing of our members in this discussion are going from here to other sites. You can judge this at other sites yourself. So don’t waste your precious time with different links.

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