[SOLVED] 程序代写代做代考 compiler interpreter midtermB-sol.dvi

midtermB-sol.dvi

Midterm Exam
CS 314, Spring ’17

March 8
Sample Solution version B

DO NOT OPEN THE EXAM
UNTIL YOU ARE TOLD TO DO SO

Name:

Rutgers ID number:

Section:

WRITE YOUR NAME ON EACH PAGE IN THE UPPER
RIGHT CORNER.

Instructions

We have tried to provide enough information to allow you to answer each of the questions. If you need
additional information, make a reasonable assumption, write down the assumption with your answer, and
answer the question. There are 6 problems, and the exam has 8 pages. Make sure that you have all pages.
The exam is worth 250 points. You have 80 minutes to answer the questions. Good luck!

This table is for grading purposes only

1 / 30

2 / 30

3 / 60

4 / 30

5 / 40

6 / 60

total / 250

1

NAME:

Problem 1 – Regular Expressions and FSAs (30 pts)

Assume that lower stands for lower case letters, i.e., {a, b, c, … z}, upper for upper case letters, i.e., {A, B,
C, … Z}, digit for digits, i.e., {0, 1, … 9}, and special for special symbols, i.e., {$, #, %, … }. The following
regular expression describes the set of valid identifiers in some programming language:

(upper|lower)special∗digit+special+

1. Give an example of a string of length 7 that

is a valid identifier in this language: A12345&

is not a valid identifier in this language: doOr123

(3 pts each)

2. Build a DFA (Deterministic Finite Automaton) for the regular expression of valid identifiers as defined
above. The start state is state 1, and the final (accepting) state is state 4. You are only allowed to
add edges with their appropriate labels, i.e., valid labels are lower, upper, digit, and special. Note that
an edge may have more than one label. (24 pts)

1 2 3 4
START

digit

digit

special

specialspecial

upper

lower

2

NAME:

Problem 2 – Context Free Grammars and Regular Expressions (30
pts)

A context-free language is a language that can be specified using a context-free grammar. A regular language
is a language that can be specified using a regular expression.

For the four languages given below, if the language is context-free, give a compact context-free grammar
in Backus-Naur-Form (BNF). If the language is regular, give a compact regular expression using the regular
expression syntax introduced in class. If a language is context-free and regular, give both specifications, a
BNF and a regular expression. You do not have to justify why you believe a language is not context-free or
not regular.

1. { a3nbn | n ≥ 0 }, with alphabet Σ = {a, b}

S ::= aaa S b |�

2. { a3nb2m | n ≥ 0, m >0 }, with alphabet Σ = {a, b}

S ::= A B

A ::= aaa A | �
B ::= bb B | bb

(aaa)∗(bb)+

3. { w1w2wR2 wR1 | w1, w2 ∈ Σ∗}, wR is w in reverse , and alphabet Σ ={a, b}

S ::= a S a | b S b | X
X ::= a X a | b X b | �

4. { w | w has at least 4 symbols}, with alphabet Σ={a, b}

S ::= A B C D X

A ::= a | b
B ::= a | b
C ::= a | b
C ::= a | b
X ::= a X | b X | �

(a | b) (a | b) (a | b) (a | b)+

3

NAME:

Problem 3 – Context Free Grammars (60 pts)

Assume the following context-free logical expression grammar in BNF over the alphabet (set of tokens)
Σ = { true, false, or, and } with the start symbol .

::= or |
::= and |
::= true | false

1. Give a left-most derivation (⇒L) for the sentence true or true and false .
⇒L
or ⇒L
or ⇒L
true or ⇒L
true or ⇒L
true or and ⇒L
true or true and ⇒L
true or true and ⇒L
true or true and false

2. Show the corresponding parse tree for your left-most deriviation.

and

true

false

or

true

3. Is the grammar LL(1)? Justify your answer using FIRST+ sets.

The grammar is not LL(1). For example, for the two rules for nonterminal , FIRST+( or
) = {true, false}, and FIRST +( ) = {true, false}. Therefore, these sets are not disjoint, which
is a requirement for the grammar to be LL(1).

4. What is the associativity of the or operator as specified in the grammar? right associative

What is the associativity of the and operator as specified in the grammar? right associative

5. Does the grammar implement the precedence of the or operator over the and operator (or binds
stronger than and)? If not, give a minimal change to the grammar (fewest number of changed rules)
such that or has precedence over and. In addition, change the grammar rules, if necessary, to make
and left associative.

4

::= and |
::= or |
::= true | false

5

NAME:

Problem 4 – Scoping (30 pts)

Assume the following program while answering the questions below.

Program A()

{ x, y, z: integer;

procedure D()

{

x = z + 2;

}

procedure C()

{

z = 3;

x = z + 4;

call D();

}

procedure B()

{ x, z: integer;

z = 1;

x = y + z;

call C();

}

// statement body of A

x = 6;

y = 7;

z = 8;

call B();

print x, y, z;

}

1. Show the output of a program execution assuming static (lexical) scoping for all variables:

x= 5 , y= 7 , z= 3

2. Show the output of a program execution assuming dynamic scoping for all variables:

x= 6 , y= 7 , z= 8

6

NAME:

Problem 5 – Pointers and Memory Allocation in C (40 pts)

int main() {
int x;
int *y;
int **z;

z = (int **) malloc (sizeof(int *));
y = (int *) malloc (sizeof(int));
x = 2;
*z = &x
*y = x;
x = x + 3;
**z = *y + 3;
printf(“x=%d, *y=%d, **z=%d
”, x, *y, **z);
return 0;

}

stack

heap

x y z

1. What is the output of the above program? (5 pts each)

x= 5 , *y= 2 , **z= 5

2. Specify, whether the following program objects are allocated on the stack (includes global variables),
on the heap, or not defined (2 pts each).

x is allocated on the stack y is allocated on the stack

z is allocated on the stack *x is allocated on the not defined

*y is allocated on the heap *z is allocated on the heap

**y is allocated on the not defined **z is allocated on the stack

3. Assume the following code segment:

int *x;
*x = 5;
printf(“%d
”, *x);

Is there a problem with this code? Assume that when you ran the code a couple of times, it printed
“5”. If you believe there is a problem, give a possible “fix” for the problem? (9 pts)

7

The content of variable x is not initialized. However, its content is used as an address of a memory
location, and that memory location is assigned the value 5.

To fix the problem, x should point to an object on the heap that is allocated as follows:

x = (int ∗) malloc(sizeof(int ∗))

This statement should be placed before statement ∗x = 5.

8

NAME:

Problem 6 – Syntax-Directed Translation (60 pts)

Assume the following logical expression grammar:

::= and |

::= true | false

instr. format description semantics

memory instructions

LOADI #true ⇒ rx load constant value #true into register rx rx ← true
LOADI #false ⇒ rx load constant value #false into register rx rx ← false

logical instructions

AND rx , ry ⇒ rz ’and’ logical operation on truth values in registers rx and ry , rz ← rx and ry
and store result into register rz

Here is a recursive descent parser that implements a compiler for the above grammar. Note that ’and’,
’true’,and ’false’ are the new tokens. Here is the important part of the code:

int expr() {

int reg, left_reg, right_reg;

switch (token) {

case ’and’: next_token();

left_reg = expr(); right_reg = expr(); reg = next_register();

CodeGen(AND, left_reg, right_reg, reg);

return reg;

case ’true’:

case ’false’: return const();

}

}

int const() {

int reg;

switch (token) {

case ’true’: next_token(); reg = next_register();

CodeGen(LOADI, true, reg);

return reg;

case ’false’: next_token(); reg = next_register();

CodeGen(LOADI, false, reg);

return reg;

}

}

Make the following assumptions:

• The value of variable “token” has been initialized correctly.
• The function CodeGen has been extended to deal with logical binary operations and logical arguments.
• The first call to function next register() the shown parser returns integer value “1”. In

other words, the first register that the generated code will be using is register r1.

• Your parser “starts” by calling function expr() on the entire input.

9

NAME:

1. Show the code that the recursive descent parser generates for input

and true and true false

will produce:

and

true and

falsetrue

loadI #true => r1

loadI #true => r2

loadI #false => r3

and r2, r3 => r4

and r1, r4 => r5

2. Change the basic recursive-descent parser to implement an interpreter for our example language. You
may insert pseudo code in the spaces marked by . You may also assume the availability of a
type boolean. No error handling is necessary. There are many possible solutions.

boolean expr() {

boolean bval1, bval2, bval;

switch (token) {

case ’and’: next_token();

bval1 = expr(); bval2 = expr();

bval = bval1 && bval2; // && is ‘‘and’’ operator

return bval;

case ’true’:

case ’false’: return const();

}

}

boolean const() {

boolean bval;

switch (token) {

case ’true’: next_token();

bval = TRUE; // TRUE is boolean constant for ‘‘true’’

return bval;

case ’false’: next_token();

bval = FALSE; // FALSE is boolean constant for ‘‘false’’

return bval;

}

}

10