csci 3120 compiler construction · 2011. 8. 28. · csci 3120 compiler construction 2nd term,...

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CSCI 3120 Compiler Construction

2nd Term, 2010/2011

Lecturer: Leung Ho Fung

Time: T9 & W5, 6

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Teacher

Leung Ho Fung Room 1011, Ho Sin-hang Engineering Building

E-mail: [email protected]

Tel.: 2609 8428

Office Hours: Tuesday and Thursday 10am – 11am

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Teaching Assistants

Li Baichuan Room 1024, HSH Engg Bldg.

[email protected]

Sheng Bin Room 913, HSH Engg Bldg. [email protected]

Tang Xingfan Room 114A, HSH Engg Bldg.

[email protected]

Wang Qiong Room 1026, HSH Engg Bldg.

[email protected]

Wang Weiming Room 1026, HSH Engg Bldg. [email protected]

Xiong Junjie Room 101, HSH Engg Bldg.

[email protected]

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Reference Book

A. V. AHO, M. S. LAM, R. SETHI AND J. D. ULLMAN, 2007. Compilers: Principles,

Techniques, and Tools: International

Edition, 2/E. Pearson Higher Education.

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Schedule:

Week Topic Aho et al.

2007 Remarks

1

Lexical

Analysis Chapter 3

2

3 Assignment 1

CHINESE NEW YEAR HOLIDAYS

4

Syntax

Analysis Chapter 4

5 Quiz 1 (15 Feb.)

6

7 Assignment 2

8 Syn-

tax-Directed

Translation

Chapter 5

(partly)

9

10 Intermedi-

ate-Code

Generation

Chapter 6

(partly)

Assignment 3

11 Quiz 2 (29 Mar.)

12 Run-time

Environment

Chapter 7

(very part-

ly)

13 Code Gen-

eration

Chapter 8

(partly)

Assignment 4

14

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Assessment Scheme:

Assignments: 40% (10% each) Quizzes: 20% (10% each)

Examination: 40%

Each assignment includes written and programming

parts.

There is no mid-term examination.

There will be two quizzes on 15 February and 29 March

(45 min each).

The final examination lasts for 2 hours.

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What Will You Learn in this Course

Understanding the phases of compilation.

Algorithms in different compiler components and their

implementations.

Understanding what a compiler can do and why it can do

it.

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What is a COMPILER?

program Hello;

begin

writeln('Hello, world!');

end.

C:\>pascal_compile hello.pas

C:\>

hello.exe

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11010101010001010001010001…

Structure of a Compiler

program Hello;begin writeln('Hello, world!');end.

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Lexical Analysis

int main(void)

{

printf("Hello, world!\n");

return 0;

}

Lexical

Analyser

int

main

(

void

)

{

printf

(

"Hello, world!\n"

⁞

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Lexical Analysis

program Hello;

begin

writeln('Hello, world!');

end.

Lexical

Analyser

program

Hello

;

begin

writeln

(

'Hello, world!'

)

⁞

;

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Lexical Analysis

with TEXT_IO;

procedure HELLO is

begin

end HELLO;

Lexical

Analyser

TEXT_IO

;

procedure

HELLO

is

begin

.

⁞

TEXT_IO

with

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Lexical Analysis

int main()

{

cout

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Tokens

i n t x 1 ; … e l s e

int

main

(

)

{

cout

⁞

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Inside tokens

= M * C ** E 2

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Inside tokens

= id 65234

* id 9745

** id 23749

832

number 43

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Inside tokens

id 9745

id 23749

832

number 43

id 65234

= M * C ** E 2

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Inside tokens

id 65234

id 9745

id 23749

832

number 43

= M * C ** E 2

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Inside tokens

id 65234

id 9745

id 23749

832

number 43

43

9745

65234

23749832

Lexeme Type …

2

…

…

C

…

…

…

M

…

…

E

…

…

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Inside tokens

id 65234

id 9745

id 23749

832

number 43

number 1

number 0

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Examples of Tokens

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Lexical Error

i n t x 1 ; … f i ( a = =

int

lexeme

;

lexeme?

x1

lexeme

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Lexical Error

i n t x 1 ; … f i ( a = =

int

lexeme

;

lexeme?

x1 The lexical ana-

lyser should think

this is an id and

return an id token.

lexeme

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Lexical Error

i n t x 1 ; … f i ~ = 1

int

lexeme

;

x1 What can we do?

lexeme

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Lexical Error

i n t x 1 ; … f i ~ = 1

int

lexeme

;

x1 ‘Panic Mode’ Recovery

Return an ‘error’ token, and

ignore all characters until

the next recognised token.

lexeme

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Lexical Error

i n t x 1 ; … f i ~ = 1

int

lexeme

;

x1 Other Possible Recovery

Delete 1 character.

Insert 1 character.

Replace 1 character.

Transpose two adjacent

characters.

lexeme

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Class Discussion

i n t 1 y 1 ;

i n t a ( ) ;

r e a l r 1 ;

i n t [ ] a ;

D O 1 0 I = 1 , 1 0

A D D 1 T O X .

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A Tricky Problem

(The for-loop in FORTRAN, called a DO-loop)

DO 10 I = 1, 10

A(I) = 0 10 CONTINUE

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A Tricky Problem


DO 10 I = 1, 25

A(I) = 0.0 10 CONTINUE

…

…

DO 10 I = 1. 25

A(I) = 0.0

10 CONTINUE

IF (A(1) .EQ. 0.0)

… …

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A Tricky Problem


DO 10 I = 1, 25


…

…

DO10I = 1.25


IF (A(1) .EQ. 0.0) … …

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Lexical Analyser and Parser

csci 3120 compiler construction · 2011. 8. 28. · csci 3120 compiler construction 2nd term,...

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