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Compiler Structures. 241-437 , Semester 1 , 2011-2012. Objective extend the expressions language compiler to generate a parse tree for the input program, and then evaluate it. 9. Creating and Evaluating a Parse Tree. Overview. 1 . The Expressions Grammar 2 . exprParse2.c

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Compiler Structures

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Compiler Structures

241-437, Semester 1, 2011-2012

  • Objective

    • extend the expressions language compiler to generate a parse tree for the input program, and then evaluate it

9. Creating and Evaluating a

Parse Tree


Overview

1. The Expressions Grammar

2. exprParse2.c

3. Parse Tree Data Structures

4. Revised Parse Functions

5.Tree Building

6.Tree Printing

7.Tree Evaluation


Source Program

In this lecture

Lexical Analyzer

Front

End

Syntax Analyzer

Semantic Analyzer

Int. Code Generator

concentrating

on parse tree

generation and

evaluation

Intermediate Code

Code Optimizer

Back

End

Target Code Generator

Target Lang. Prog.


1. The Expressions Grammar

  • It's LL(1) grammar:

    Stats => ( [ Stat ] \n )*

    Stat => let ID = Expr | Expr

    Expr => Term ( (+ | - ) Term )*

    Term => Fact ( (* | / ) Fact ) *

    Fact => '(' Expr ')' | Int | Id


An Expressions Program (test3.txt)

5 + 6

let x = 2

3 + ( (x*y)/2) // comments

// y

let x = 5

let y = x /0

// comments


exprParse2.c

  • A recursive descent parser using the expressions language.

  • This version of the parser differs from exprParse1.c by having the parse functions (e.g. statements(), statement()) create a parse tree as they execute.

continued


  • There's a new printTree() function which prints the final tree, and evalTree() which evaluates it.

  • Usage:

    $ gcc -Wall -o exprParse2 exprParse2.c$ ./exprParse2 < test1.txt


Output for test1.txt

let x = 2

let y = 3 + x

> exprParse2 < test1.txt

\n

\n

NULL

=

x

2

=

y

+

3

x

\n

\n

=

printed

tree;

same as

y

+

=

NULL

x

2

3

x

continued


x being declared

x = 2

== 2

y being declared

y = 5

== 5

>

evaluation of the

parse tree


3. Parse Tree Data Structures

typedef struct TreeNode {

Token operTok;

union {

char *id;

int value;

struct {struct TreeNode *left, *right;} branches;

} u;

} Tree;

A tree is made from TreeNodes.


Graphically

one of ID, INT,NEWLINE,ASSIGNOP,PLUSOP, MINUSOP,

MULTOP, DIVOP

TreeNode

operTok

id

variable name (for ID)

OR

a union, u

value

integer (for INT)

OR

children pointers of

this node (used by

NEWLINE, ASSIGNOP,PLUSOP, MINUSOP,

MULTOP, DIVOP)

left

right

branches


Macros for Using TreeNode Fields

#define TreeOper(t) ((t)->operTok)

#define TreeID(t) ((t)->u.id)

#define TreeValue(t) ((t)->u.value)

#define TreeLeft(t) ((t)->u.branches.left)

#define TreeRight(t) ((t)->u.branches.right)


4. Revised Parse Functions

  • The parse functions have the same 'shape' as the ones in exprParse0.c, but now call tree building functions, and return a Tree result.

  • Functions:

    • main(), statements(), statement(), expression(), term(), factor()


int main(void)

{

nextToken();

statements();

match(SCANEOF);

return 0;

}

int main(void)

// parse, then print and evaluate the resulting tree

{

Tree *t;

nextToken();

t = statements();

match(SCANEOF);

printTree(t, 0);

printf("\n\n");

evalTree(t);

return 0;

}

main() Before and After


statements() Before and After

with no semantic actions

void statements(void)

// statements ::= { [ statement] '\n' }

{

dprint("Parsing statements\n");

while (currToken != SCANEOF) {

if (currToken != NEWLINE)

statement();

match(NEWLINE);

}

} // end of statements()


Tree *statements(void)

{

Tree *t, *left, *statTree;

left = NULL;

dprint("Parsing statements\n");

while (currToken != SCANEOF) {

if (currToken != NEWLINE)

statTree = statement();

else

statTree = NULL;

match(NEWLINE);

if (statTree != NULL) {

t = makeTreeNode(NEWLINE, left, statTree);

left = t;

}

}

return left;

} // end of statements()


Tree Structure for statements

  • A statements sequence:

    s1 \n1 s2 \n2 s3 \n3

    becomes:

\n3

\n2

s3

s2

\n1

s1

NULL


statement() Before and After

with no semantic actions

void statement(void)

// statement ::= ( 'let' ID '=' EXPR ) | EXPR

{

if (currToken == LET) {

match(LET);

match(ID);

match(ASSIGNOP);

expression();

}

else

expression();

} // end of statement()


Tree *statement(void)

{

Tree *t, *idTree, *exprTree;

dprint("Parsing statement\n");

if (currToken == LET) {

match(LET);

idTree = matchId();

// build tree node, not symbol table entry

match(ASSIGNOP);

exprTree = expression();

t = makeTreeNode(ASSIGNOP, idTree, exprTree);

}

else // expression

t = expression();

return t;

} // end of statement()


Tree Structures for statement

=

or

expr

tree

ID

node

expr

tree


expression() Before and After

with no semantic actions

void expression(void)

// expression ::= term ( ('+'|'-') term )*

{

term();

while((currToken == PLUSOP) || (currToken == MINUSOP)) {

match(currToken);

term();

}

} // end of expression()


Tree *expression(void)

{ Tree *t, *left, *right;

int isAddOp;

dprint("Parsing expression\n");

left = term();

while((currToken == PLUSOP)||(currToken == MINUSOP)) {

isAddOp = (currToken == PLUSOP) ? 1 : 0;

nextToken();

right = term();

if (isAddOp == 1) // addition

t = makeTreeNode(PLUSOP, left, right);

else // subtraction

t = makeTreeNode(MINUSOP, left, right);

left = t;

}

return left;

} // end of expression()


Tree Structure for expression

  • An expression sequence:

    t1 +1 t2 - t3 +2 t4

    becomes:

+2

-

t4

t3

+1

t2

t1


term() Before and After

with no semantic actions

void term(void)

// term ::= factor ( ('*'|'/') factor )*

{

factor();

while((currToken == MULTOP) || (currToken == DIVOP)) {

match(currToken);

factor();

}

} // end of term()


Tree *term(void)

{ Tree *t, *left, *right;

int isMultOp;

dprint("Parsing term\n");

left = factor();

while((currToken == MULTOP) || (currToken == DIVOP)) {

isMultOp = (currToken == MULTOP) ? 1 : 0;

nextToken();

right = factor();

if (isMultOp == 1) // multiplication

t = makeTreeNode(MULTOP, left, right);

else // division

t = makeTreeNode(DIVOP, left, right);

left = t;

}

return left;

} // end of term()


Tree Structure for term

  • An term sequence:

    f1 *1 f2 / f3 *2 f4

    becomes:

*2

/

f4

f3

*1

f2

f1


factor() Before and After

with no semantic actions

void factor(void)

// factor ::= '(' expression ')' | INT | ID

{

if(currToken == LPAREN) {

match(LPAREN);

expression();

match(RPAREN);

}

else if(currToken == INT)

match(INT);

else if (currToken == ID)

match(ID);

else

syntax_error(currToken);

} // end of factor()


Tree *factor(void)

{ Tree *t = NULL;

dprint("Parsing factor\n");

if(currToken == LPAREN) {

match(LPAREN);

t = expression();

match(RPAREN);

}

else if(currToken == INT) {

t = makeIntLeaf(currTokValue);

match(INT);

}

else if (currToken == ID)

t = matchId();// do not access symbol table

else

syntax_error(currToken);

return t;

} // end of factor()


Match an ID (Extended)

Tree *matchId(void)

{

Tree *t;

if (currToken == ID)

t = makeIDLeaf(tokString);

match(ID);

return t;

} // end of matchID()


Tree Structure for factor

  • There are three possible nodes:

tree

node

INT

node

ID

node

or

or


5. Tree Building

TreeNode

  • The nodes in a parse tree are connected by the parse functions.

  • A tree node can have three different shapes:

operTok

id

OR

value

a union

OR

left

right

branches


Making a Tree Node

Tree *treeMalloc(void)

// a tree node with no fields specified

{

Tree *t;

t = (Tree *) malloc( sizeof(Tree) );

if(t == NULL) {/* out of memory? */

perror("Tree Node not made; out of memory");

exit(1);

}

return t;

} // end of treeMalloc()


Making an ID Node

operTok

ID

"id str"

id

Tree *makeIDLeaf(char *id)

{

Tree *t;

t = treeMalloc();

TreeOper(t) = ID;

TreeID(t) = (char *) malloc(strlen(id)+1);

strcpy(TreeID(t), id);

return t;

} // end of makeIDLeaf()

no symbol table entry

created yet


Making an INT Node

operTok

INT

Tree *makeIntLeaf(int value)

{

Tree *t;

t = treeMalloc();

TreeOper(t) = INT;

TreeValue(t) = value;

return t;

} // end of makeIntLeaf()

integer

value


Making a Node with Children

Tree *makeTreeNode(Token op, Tree *left, Tree *right)

/* Build an internal tree node, which contains an operator and points to two subtrees.*/

{

Tree *t;

t = treeMalloc();

TreeOper(t) = op;

TreeLeft(t) = left;

TreeRight(t) = right;

return t;

} // end of makeTreeNode()

operTok

op

branches

left

right


6. Tree Printing

  • The printTree() function recurses over the tree, and does three different things depending on the three possible 'shapes' for a tree node.

  • It includes an indent counter, which is used to print spaces (indents) in front of the node information.


void printTree(Tree *t, int indent)

// print a tree, indenting by indent spaces

{

printIndent(indent);

if (t == NULL) {

printf("NULL\n");

return;

} :

continued


Token tok = TreeOper(t);

if (tok == INT)

printf("%d\n", TreeValue(t));

else if (tok == ID)

printf("%s\n", TreeID(t));

else { // operator

if (tok == NEWLINE)

printf("\\n\n"); // show the \n

else

printf("%s\n", tokSyms[tok]);

printTree(TreeLeft(t), indent+2);

printTree(TreeRight(t), indent+2);

}

} // end of printTree()


void printIndent(int n)

{ int spaces;

for(spaces = 0; spaces != n; spaces++)

putchar(' ');

} // end of printIndent()


> exprParse2 < test2.txt

\n

\n

\n

NULL

=

x56

2

=

bing_BONG

-

*

27

2

x56

*

5

/

67

3

Tree Printing Examples

let x56 = 2

let bing_BONG = (27 * 2) - x56

5 * (67 / 3)


Graphically

\n

\n

*

/

5

\n

=

67

3

bing_BONG

-

S3

=

NULL

x56

*

x56

2

27

2

S1

S2


test3.txt

5 + 6

let x = 2

3 + ( (x*y)/2) // comments

// y

let x = 5

let y = x /0

// comments


> exprParse2 < test3.txt

\n

\n

\n

\n

\n

NULL

+

5

6

=

x

2

+

3

/

*

x

y

2

=

x

5

=

y

/

x

0


7. Tree Evaluation

  • Tree evaluation works in two stages:

    • evalTree() searches over the tree looking for subtrees which start with an operator which is not NEWLINE

    • these subtrees are evaluated by eval(), using the operators in their nodes


Finding non-NEWLINEs

\n

evalTree()

used here

\n

*

/

5

\n

=

67

3

bing_BONG

-

=

NULL

x56

*

x56

2

27

2

eval() used here


Code

void evalTree(Tree *t)

{

if (t == NULL)

return;

Token tok = TreeOper(t);

if (tok == NEWLINE) {

evalTree( TreeLeft(t));

evalTree( TreeRight(t));

}

else

printf("== %d\n", eval(t));

} // end of evalTree()


The operator can

be one of ID, INT,ASSIGNOP,PLUSOP, MINUSOP,

MULTOP, DIVOP

int eval(Tree *t)

{

SymbolInfo *si;

if (t == NULL)

return 0;

Token tok = TreeOper(t);

if (tok == ID) {

si = getIDEntry( TreeID(t) ); // lookup ID in symbol table

return si->value;

} :

7 possibilities

continued


else if (tok == INT)

return TreeValue(t);

else if (tok == ASSIGNOP) { // id = expr

si = evalID(TreeLeft(t)); //add ID to sym. table

int result = eval(TreeRight(t));

si->value = result;

printf("%s = %d\n", si->id, result);

return result;

}

else if (tok == PLUSOP)

return eval(TreeLeft(t)) + eval(TreeRight(t));

else if (tok == MINUSOP)

return eval(TreeLeft(t)) - eval(TreeRight(t));

:


else if (tok == MULTOP)

return eval(TreeLeft(t)) * eval(TreeRight(t));

else if (tok == DIVOP) {

int right = eval(TreeRight(t));

if (right == 0) {

printf("Error: Div by 0; using 1 instead\n");

return eval(TreeLeft(t));

}

else

return eval(TreeLeft(t)) / right;

}

return 0; // shouldn't reach here

} // end of eval()


SymbolInfo *evalID(Tree *t)

{

char *id = TreeID(t);

return getIDEntry(id); // create sym. table entry for id

} // end of evalID()

this function finds or creates a

symbol table entry for the id, and

return a pointer to the entry

(same as in exprParse1.c)


Evaluation Examples

let x = 2

let y = 3 + x

$ ./exprParse2 < test1.txt

:

x declared

x = 2

== 2

y declared

y = 5

== 5


$ ./exprParse2 < test2.txt

:

x56 declared

x56 = 2

== 2

bing_BONG declared

bing_BONG = 52

== 52

== 110

// test2.txt example

let x56 = 2

let bing_BONG = (27 * 2) - x56

5 * (67 / 3)


5 + 6

let x = 2

3 + ( (x*y)/2) // comments

// y

let x = 5

let y = x /0

$ ./exprParse2 < test3.txt

:

== 11

x declared

x = 2

== 2

y declared

== 3

x = 5

== 5

Error: Division by zero; using 1 instead

y = 5

== 5


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