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Lexical and Syntax Analysis(of Programming Languages)
Abstract Syntax
Lexical and Syntax Analysis(of Programming Languages)
Abstract Syntax
What is Parsing?
String ofcharacters
Easy for humansto write
Easy for programsto process
Parser
A parser also checks that the input stringis well-formed, and if not, rejects it.
Data structure
What is Parsing?
String ofcharacters
Easy for humansto write
Easy for programsto process
Parser
A parser also checks that the input stringis well-formed, and if not, rejects it.
Data structure
Example 1
Charlton, 49
Lineker, 48
Beckham, 17
CSV (Comma Separated Value)
Array of pairs
Parser
“Charlton”
49
“Lineker”
48
“Beckham”
17
Example 1
Charlton, 49
Lineker, 48
Beckham, 17
CSV (Comma Separated Value)
Array of pairs
Parser
“Charlton”
49
“Lineker”
48
“Beckham”
17
Concrete andAbstract Syntax
The concrete syntax is a set ofrules that describe valid inputsto the parser.
The abstract syntax is a set ofrules that describe valid outputsfrom the parser.
The data structure produced bya parser is commonly termedthe abstract syntax tree.
Concrete andAbstract Syntax
The concrete syntax is a set ofrules that describe valid inputsto the parser.
The abstract syntax is a set ofrules that describe valid outputsfrom the parser.
The data structure produced bya parser is commonly termedthe abstract syntax tree.
Concrete andAbstract Syntax
String ofcharacters
Conforms to theConcrete Syntaxof the language
Conforms to theAbstract Syntaxof the language
Parser
Data structure
Concrete andAbstract Syntax
String ofcharacters
Conforms to theConcrete Syntaxof the language
Conforms to theAbstract Syntaxof the language
Parser
Data structure
Abstract syntax
The abstract syntax is usuallyspecified as a data type in theprogramming language beingused, in our case C. Example:
typedef struct {char* name;int goals;
} Player;
typedef struct {Player* players;int size;
} Squad;
An abstract syntax tree is avalue of this type.
Abstract syntax
The abstract syntax is usuallyspecified as a data type in theprogramming language beingused, in our case C. Example:
typedef struct {char* name;int goals;
} Player;
typedef struct {Player* players;int size;
} Squad;
An abstract syntax tree is avalue of this type.
This Chapter
How:
to define the abstract syntax
to construct abstract syntax trees
in the programming language C.
Also revisits some important Cprogramming techniques.
If you need a C tutorial then thefollowing books are recommended.
This Chapter
How:
to define the abstract syntax
to construct abstract syntax trees
in the programming language C.
Also revisits some important Cprogramming techniques.
If you need a C tutorial then thefollowing books are recommended.
POINTERS
Pointer: a variable that holds the address of a core storage location. [The Free Dictionary]
POINTERS
Pointer: a variable that holds the address of a core storage location. [The Free Dictionary]
Pointers
Declare a variable x of type intand initialise it to the value 10.
int x = 10; 10
x:
int* p;p:
p = &x;p:
10
x:
Declare a variable p of type int*(read: int pointer).
Make p point to x (or assign theaddress of x to p).
Pointers
Declare a variable x of type intand initialise it to the value 10.
int x = 10; 10
x:
int* p;p:
p = &x;p:
10
x:
Declare a variable p of type int*(read: int pointer).
Make p point to x (or assign theaddress of x to p).
Pointers
Print p (here, the address of x).
printf("%i\n", p );p:
10
x:
Print the value pointed to by p(here, the value of x).
printf("%i\n", *p );p:
10
x:
Assign 20 to the location pointedto by p.
*p = 20;p:
20
x:
Pointers
Print p (here, the address of x).
printf("%i\n", p );p:
10
x:
Print the value pointed to by p(here, the value of x).
printf("%i\n", *p );p:
10
x:
Assign 20 to the location pointedto by p.
*p = 20;p:
20
x:
Exercise 1
void swap(int* x, int* y) {
int tmp;tmp = *x;*x = *y;*y = tmp;
}
void main(){
int a = 1;int b = 2;swap(&a, &b);printf("a=%i, b=%i\n", a, b);
}
What is printed by the followingprogram?
Exercise 1
void swap(int* x, int* y) {
int tmp;tmp = *x;*x = *y;*y = tmp;
}
void main(){
int a = 1;int b = 2;swap(&a, &b);printf("a=%i, b=%i\n", a, b);
}
What is printed by the followingprogram?
DYNAMIC ALLOCATION
Dynamic Allocation: the allocation of memory storage for use in a computer program. [The Free Dictionary]
DYNAMIC ALLOCATION
Dynamic Allocation: the allocation of memory storage for use in a computer program. [The Free Dictionary]
Array allocation
int* p;p:
p = malloc(4 * sizeof(int));
p:
Declare a variable p of type int*.
Allocate memory for an array of4 int values and let point p to it.
Array allocation
int* p;p:
p = malloc(4 * sizeof(int));
p:
Declare a variable p of type int*.
Allocate memory for an array of4 int values and let point p to it.
Array indexing
*p = 10;p:
10
Assign 10 to the location pointedto by p.
p[0] = 20;p:
20
Assign 20 to the first element ofthe array pointed to by p.
p[2] = p[0];p:
20 20
Copy the first element of thearray to the third element.
Array indexing
*p = 10;p:
10
Assign 10 to the location pointedto by p.
p[0] = 20;p:
20
Assign 20 to the first element ofthe array pointed to by p.
p[2] = p[0];p:
20 20
Copy the first element of thearray to the third element.
Array deallocation
free(p);p:
When finished with an arrayallocated by malloc, call free torelease the space, otherwiseyour program may run out ofmemory.
Space released, so it canbe reused by future callsto malloc.
Array deallocation
free(p);p:
When finished with an arrayallocated by malloc, call free torelease the space, otherwiseyour program may run out ofmemory.
Space released, so it canbe reused by future callsto malloc.
STRINGS
String: a series of consecutive characters. [The Free Dictionary]
STRINGS
String: a series of consecutive characters. [The Free Dictionary]
Strings
char* s = “hi”;s:
h i \0
Declare a variable s, initialised topoint to the string “hi”.
s = s + 1;
Let s point to the next character.
s = s - 1;
And let s point to the previouscharacter again.
s:
h i \0
s:
h i \0
Strings
char* s = “hi”;s:
h i \0
Declare a variable s, initialised topoint to the string “hi”.
s = s + 1;
Let s point to the next character.
s = s - 1;
And let s point to the previouscharacter again.
s:
h i \0
s:
h i \0
Exercise 2
int f(char* s){
int i = 0;while (s[i] != '\0') i++;return i;
}
void main(){
char* x = “Hello”;printf(“%i\n”, f(x));
}
What is printed by the followingprogram?
Exercise 2
int f(char* s){
int i = 0;while (s[i] != '\0') i++;return i;
}
void main(){
char* x = “Hello”;printf(“%i\n”, f(x));
}
What is printed by the followingprogram?
USER-DEFINED TYPES
Type: the general character or structure held in common by a number of things. [The Free Dictionary]
USER-DEFINED TYPES
Type: the general character or structure held in common by a number of things. [The Free Dictionary]
Type definitions
typedef int Integer;
A typedef declaration allows anew name to be given to a type.
typedef char* String;
A new nameExisting type
String s; /* Declare a string s */Integer i; /* and an integer i */i = 0;s = “hello”;
Example use:
Type definitions
typedef int Integer;
A typedef declaration allows anew name to be given to a type.
typedef char* String;
A new nameExisting type
String s; /* Declare a string s */Integer i; /* and an integer i */i = 0;s = “hello”;
Example use:
Enumerations
enum colour {RED, GREEN, BLUE};
An enum declaration introducesa new type whose values aremembers of a given set.
Example use:
New type Possible values
enum colour c;c = RED;if (c == RED) printf(“Red\n”);
typedef enum colour Colour;
Give it a shorter name:
Enumerations
enum colour {RED, GREEN, BLUE};
An enum declaration introducesa new type whose values aremembers of a given set.
Example use:
New type Possible values
enum colour c;c = RED;if (c == RED) printf(“Red\n”);
typedef enum colour Colour;
Give it a shorter name:
Structures
An struct declaration introducesa new type that is a conjunctionof one or more existing types.
struct rectangle {float width;float height;
};
A width anda height
New type
struct rectangle r;r.width = 10;r.height = 20;
Example use:
struct circle {float radius;
}
A circle:
Structures
An struct declaration introducesa new type that is a conjunctionof one or more existing types.
struct rectangle {float width;float height;
};
A width anda height
New type
struct rectangle r;r.width = 10;r.height = 20;
Example use:
struct circle {float radius;
}
A circle:
Unions
An union declaration introducesa new type that is a disjunctionof one or more existing types.
union shape {struct circle circ;struct rectangle rect;
};
A circle ora rectangle
New type
struct shape s;s.circ.radius = 10;
Example use:
Unions
An union declaration introducesa new type that is a disjunctionof one or more existing types.
union shape {struct circle circ;struct rectangle rect;
};
A circle ora rectangle
New type
struct shape s;s.circ.radius = 10;
Example use:
Tagged unions
Often a tag is used to denote theactive disjunct of a union.
Another definition of shape:
struct shape {enum { CIRCLE, RECTANGLE } tag;union {
struct circle circ;struct rectangle rect;
};};
Tagged unions
Often a tag is used to denote theactive disjunct of a union.
Another definition of shape:
struct shape {enum { CIRCLE, RECTANGLE } tag;union {
struct circle circ;struct rectangle rect;
};};
Tagged unions
struct shape s, t;
s.tag = CIRCLE;s.circ.radius = 10;
t.tag = RECTANGLE;t.rect.width = 5;t.rect.height = 15;
Example: s is a circle and t is arectangle, and both are of typestruct shape.
Tagged unions
struct shape s, t;
s.tag = CIRCLE;s.circ.radius = 10;
t.tag = RECTANGLE;t.rect.width = 5;t.rect.height = 15;
Example: s is a circle and t is arectangle, and both are of typestruct shape.
Tagged unions
Example: compute the area ofany given shape s.
float area(struct shape s){
if (s.tag == CIRCLE) {float r = s.circ.radius;return (3.14 * r * r);
}if (s.tag == RECTANGLE) {
return (s.rect.width *s.rect.height);
}}
Tagged unions
Example: compute the area ofany given shape s.
float area(struct shape s){
if (s.tag == CIRCLE) {float r = s.circ.radius;return (3.14 * r * r);
}if (s.tag == RECTANGLE) {
return (s.rect.width *s.rect.height);
}}
Recursive structures
A value of type struct t maycontain a value of type struct t*.
struct list {int head;struct list* tail;
};
typedef struct list List;
(*xs).head ≡ xs->head
Suppose x is a value of type List*.
(*(*xs).tail).head ≡ xs->tail->head
Recursive structures
A value of type struct t maycontain a value of type struct t*.
struct list {int head;struct list* tail;
};
typedef struct list List;
(*xs).head ≡ xs->head
Suppose x is a value of type List*.
(*(*xs).tail).head ≡ xs->tail->head
Recursive structures
Example: inserting an item ontothe front of a linked list.
List* insert(List* xs, int x){
List* ys = malloc(sizeof(List));ys->head = x;ys->tail = xs;return ys;
}
Recursive structures
Example: inserting an item ontothe front of a linked list.
List* insert(List* xs, int x){
List* ys = malloc(sizeof(List));ys->head = x;ys->tail = xs;return ys;
}
CASE STUDY
A simplifier for arithmetic expressions.
CASE STUDY
A simplifier for arithmetic expressions.
Concrete syntax
Consider the following concretesyntax for arithmetic expressions,where v ranges over variablenames and n over integers.
e → v| n| e + e| e * e| ( e )
Example expression:
x * y + (z * 10)
Concrete syntax
Consider the following concretesyntax for arithmetic expressions,where v ranges over variablenames and n over integers.
e → v| n| e + e| e * e| ( e )
Example expression:
x * y + (z * 10)
Simplification
Consider the algebraic law:
∀x. x * 1 = x
Example simplification:
x * (y * 1) → x * y
This law can be used to simplifyexpressions by using it as arewrite rule from left to right.
Simplification
Consider the algebraic law:
∀x. x * 1 = x
Example simplification:
x * (y * 1) → x * y
This law can be used to simplifyexpressions by using it as arewrite rule from left to right.
Problem
1. Define an abstract syntax, in C,for arithmetic expressions.
2. Show how to construct abstractsyntax trees that representarithmetic expressions.
3. Implement the simplification as aC function that takes and returnsan abstract syntax tree.
Problem
1. Define an abstract syntax, in C,for arithmetic expressions.
2. Show how to construct abstractsyntax trees that representarithmetic expressions.
3. Implement the simplification as aC function that takes and returnsan abstract syntax tree.
Abstract syntax
typedef enum { ADD, MUL } Op;
struct expr {enum { VAR, NUM, APP } tag;union {
char* var;int num;struct {
struct expr* e1;Op op;struct expr* e2;
} app;};
};
typedef struct expr Expr;
A variable ora number oran op and twosub-expressions
Abstract syntax
typedef enum { ADD, MUL } Op;
struct expr {enum { VAR, NUM, APP } tag;union {
char* var;int num;struct {
struct expr* e1;Op op;struct expr* e2;
} app;};
};
typedef struct expr Expr;
A variable ora number oran op and twosub-expressions
Constructors
Expr* mkVar(char* v) {Expr* e = malloc(sizeof(Expr));e->tag = VAR; e->var = v;return e;
}
Expr* mkNum(int n) {Expr* e = malloc(sizeof(Expr));e->tag = NUM; e->num = n;return e;
}
Expr* mkApp(Expr* e1, Op op, Expr* e2) {Expr* e = malloc(sizeof(Expr));e->tag = APP; e->app.op = op;e->app.e1 = e1; e->app.e2 = e2;return e;
}
Constructors
Expr* mkVar(char* v) {Expr* e = malloc(sizeof(Expr));e->tag = VAR; e->var = v;return e;
}
Expr* mkNum(int n) {Expr* e = malloc(sizeof(Expr));e->tag = NUM; e->num = n;return e;
}
Expr* mkApp(Expr* e1, Op op, Expr* e2) {Expr* e = malloc(sizeof(Expr));e->tag = APP; e->app.op = op;e->app.e1 = e1; e->app.e2 = e2;return e;
}
Abstract syntax trees
mkApp( mkVar("x"), ADD, mkApp( mkVar("y")
, MUL, mkNum(2)))
An abstract syntax tree thatrepresents the expression
x + y * 2
can be constructed by thefollowing C expression
Abstract syntax trees
mkApp( mkVar("x"), ADD, mkApp( mkVar("y")
, MUL, mkNum(2)))
An abstract syntax tree thatrepresents the expression
x + y * 2
can be constructed by thefollowing C expression
Simplification
void simplify(Expr* e){
if (e->tag == APP&& e->app.op == MUL&& e->app.e2->tag == NUM&& e->app.e2->num == 1) {
*e = *(e->app.e1);}if (e->tag == APP) {
simplify(e->app.e1);simplify(e->app.e2);
}}
∀x. x * 1 = x
is implemented by
Simplification
void simplify(Expr* e){
if (e->tag == APP&& e->app.op == MUL&& e->app.e2->tag == NUM&& e->app.e2->num == 1) {
*e = *(e->app.e1);}if (e->tag == APP) {
simplify(e->app.e1);simplify(e->app.e2);
}}
∀x. x * 1 = x
is implemented by
Homework exercises
Extend the simplifier to exploitthe following algebraic law.
Implement a pretty printer thatprints an abstract syntax tree ina concrete form.
void print(Expr* e){
...}
∀x. x * 0 = 0
Homework exercises
Extend the simplifier to exploitthe following algebraic law.
Implement a pretty printer thatprints an abstract syntax tree ina concrete form.
void print(Expr* e){
...}
∀x. x * 0 = 0
Motivation for LSA
In LSA, we are interested in howto implement the following kindof function
Expr* parse(char* string){
...}
It takes a string conforming tothe concrete syntax and returnsan abstract syntax tree.
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