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CS6612 COMPILER LABORATORY SYLLABUS 1. Implementation of Symbol Table 2. Develop a lexical analyzer to recognize a few patterns in C. (Ex. identifiers, constants, comments, operators etc.) 3. Implementation of Lexical Analyzer using Lex Tool 4. Generate YACC specification for a few syntactic categories. a) Program to recognize a valid arithmetic expression that uses operator +, - , * and /. b) Program to recognize a valid variable which starts with a letter followed by any number of letters or digits. c) Implementation of Calculator using LEX and YACC 5. Convert the BNF rules into Yacc form and write code to generate Abstract Syntax Tree. 6. Implement type checking 7. Implement control flow analysis and Data flow Analysis 8. Implement any one storage allocation strategies(Heap,Stack,Static) 9. Construction of DAG 10. Implement the back end of the compiler which takes the three address code and produces the 8086 assembly language instructions that can be assembled and run using a 8086 assembler. The target assembly instructions can be simple move, add, sub, jump. Also simple addressing modes are used. 11. Implementation of Simple Code Optimization Techniques




COMPILER LAB

EXP.NO.

NAME OF THE EXPERIMENT

1

Implementation of Symbol Table

2

Develop a lexical analyzer to recognize a few patterns in C.

3

Implementation of Lexical Analyzer using Lex Tool

4

Generate YACC specification for a few syntactic categories. a) Program to recognize a valid arithmetic expression that uses operator +, - , * and /. b) Program to recognize a valid variable var iable which starts start s with a letter followed by any number of letters or digits. c) Implementation of Calculator using LEX and YACC

5

Convert the BNF rules into Yacc form and write code to generate Abstract Syntax Tree.

6

Implement type checking

7

Implement stack storage allocation

8

Construction of DAG

9

Implement the back end of the compiler

10

Implement the front end of the compiler

11

Implementation of Simple Code Optimization Techniques. Techniques .




Compiler is a System Software that converts High level language to low level language .We human beings can‟t program in machine language (low level lang.) understood by Computers so so we program in high level language and compiler is the software which bridges the gap between user and computer. It‟s a very complicated piece of software which took 18 man years to build first compiler. To build this S/w it is is divided into six phases which are 1) Lexical Analysis 2)Syntax Analysis 3)Semantic Analysis 4)Intermediate Code 5)Code Optimization 6)Code Generation. Lex helps write programs whose control flow is directed by instances of regular expressions in the input stream. It is well suited for editor-script type transformations and for segmenting input in preparation for a parsing routine. Lex source is a table of regular expressions and corresponding program fragments. The table is translated to a program which reads an input stream, copying it to an output stream and partitioning the input into strings which whic h match the t he given expressions. As each eac h such su ch string is recognized the corresponding program fragment is executed. The recognition of the expressions is performed by a deterministic finite automaton generated by Lex. The program fragments written by the user are executed in the order in which the corresponding regular expressions occur in the input stream. The lexical analysis programs written with Lex accept ambiguous specifications and choose the longest match possible at each input point. If necessary, substantial lookahead is performed on the input, but the input stream will be backed up to the end of the current partition, so that the t he user has general freedom to manipulate it. List of Apparatus: Hardware Requirements: Processsor: Pentium I RAM: 128MB Hard Disk 40 GB Floppy Drive 1.44MB Software Requirements: Language: C/C++




EXERCISE – 1 1 SYMBOL TABLE IMPLEMENTATION Aim: To write a C program to implement a symbol table. Steps : 1) Start the program. 2) Get the input input from the user w ith the terminating symbol „$‟. 3) Allocate memory memory for the variable by dyna mic memory memory allocation function. 4) If the next character of the symbol is an operator then only the memory is allocated. 5) While reading, read ing, the input symbol is inserted inser ted into symbol table tab le along with its memoryaddress. 6) The steps are repeated till „$‟ is reached. 7) To reach a variable, enter the variable to the searched and symbol table has be e n c he c ke d for corresponding correspo nding variable, the variable along with its address is displayed d isplayed as result. 8)Stop the program. Program: #include #include #include #include #include #include void main() { int i=0,j=0,x=0,n,flag=0; void *p,*add[5]; char ch,srch,b[15],d[15],c; clrscr(); printf("Expression terminated ter minated by $ : "); while((c=getchar())!='$') { b[i]=c;i++; } n=i-1; printf("Given Expression : ");



i=0; while(i<=n) { printf("%c",b[i]); i++; } printf("\nSymbolTable\n"); printf("\nSymbolTable\n "); printf("Symbol\taddr\ttype"); printf("Symbol\tad dr\ttype"); while(j<=n) { c=b[j]; if(isalpha(toascii(c))) { if(j==n) { p=malloc(c); add[x]=p; d[x]=c; printf("%c\t%d\tidentifier",c,p);; printf("%c\t%d\tidentifier",c,p) }else { ch=b[j+1]; if(ch=='+'||ch=='-'||ch=='*'||ch=='=') { p=malloc(c); add[x]=p; d[x]=c; printf("\n%c\t%d\tidentifier\n",c,p); printf("\n%c\t%d \tidentifier\n",c,p); x++; }}} j++; } printf("\nThe symbol symbo l is to be searched"); srch=getch(); for(i=0;i<=x;i++) { if(srch==d[i]) { printf("\nSymbol Found"); Found") ; printf("\n%c%s%d\n",srch," printf("\n%c%s%d\n",src h," @address ",add[i]); flag=1; }} if(flag==0) printf("\nSymbol Not Found"); getch(); }



OUTPUT






Result Thus the above the program is executed and the required output is obtained. EXERCISE – 2 2

IMPLEMENTATION OF LEXICAL ANALYZER IN C Aim: To write a C Program to implement a Lexical analyzer. Steps: 1)Start the program. 2)Declare all the variables and file pointers. 3 ) D i s p l a y t h e i n p u t p r o g r a m. m. 4)Separate the keyword in the program and display it. 5)Display the header files of the input program. 6)Separate the operators of the input program and display it. 7 ) P r i nt nt t h e p u n c t u a t i o n m a r k s . 8)Print the constant that are present in input program. 9)Print the identifiers of the input program. 10)Stop the program. Program: #include #include #include #include void main() { FILE *fp; int i,j; char arr[100],k; char kw[10][10]={"int","float","double","end","main","void","include","printf","scanf"}; char hf[2][10]={"stdio.h","conio.h"}; char op[5]={'+','-','*','/','%'}; char punc[6]={'(',')','{','}',','}; clrscr(); fp=fopen("input.c","r"); printf("InputProgram\n"); while(!feof(fp))


{ arr[0]=fgetc(fp); printf("%c",arr[0]); } fclose(fp); printf("\nSymboltable\n"); printf("\nSymboltab le\n"); fp=fopen("input.c","r"); printf("\nKeywords"); while(!feof(fp)) { arr[0]=fgetc(fp); fscanf(fp,"%s",arr); for(i=0;i<10;i++) { if(strcmp(arr,kw[i])==0) { printf("\t%s",arr); }}} fclose(fp); fp=fopen("input.c","r"); printf("\nHeader files"); while(!feof(fp)) { arr[0]=fgetc(fp); fscanf(fp,"%s",arr); for(i=0;i<2;i++) { if(strcmp(arr,hf[i])==0) { printf("\t%s",arr); printf("\t%s",arr) ; }}} fclose(fp); fp=fopen("input.c","r"); printf("\nOperato rs"); while(!feof(fp)) { arr[0]=fgetc(fp); for(i=0;i<5;i++) { if(arr[0]==op[i]) { printf("\t%c",arr[0]); printf("\t%c",arr[0]) ; }}} fclose(fp); fp=fopen("input.c","r"); printf("\npunctuation"); printf("\npunctu ation");




while(!feof(fp)) { arr[0]=fgetc(fp); for(i=0;i<6;i++) { if(arr[0]==punc[i]) { printf("\t%c",arr[0]);}}} printf("\t%c",arr[0]) ;}}} fclose(fp); fp=fopen("input.c","r"); printf("\nConstants"); printf("\nConstant s"); while(!feof(fp)) { arr[0]=fgetc(fp); if(isdigit(arr[0])) { printf(" %c ",arr[0]);}} fclose(fp); fp=fopen("input.c","r"); printf("\nidentifier printf("\nident ifier "); while(!feof(fp)) { fscanf(fp,"%s",arr); for(i=0;i<2;i++) { if(strcmp(arr,kw[i])==0) { fscanf(fp,"%s",arr); j=0; while(j
INPUT: #include #include void main() { int a,b,c; a=10;




b=5; c=a+b; printf(“The sum is %d”,c); getch(); }

OUTPUT:






Result:

Thus the above the program is executed and the required output is obtained EXERCISE – 3 3 IMPLEMENTATION IMPLEMENTATION OF LEXICAL ANALYSER USING LEX TOOL Aim: To write a lex program to implement the lexical analyzer. Steps: 1.Start the program 2.Open a file file.c in read and include the yylex() tool for input scanning. 3 . D e f i ne n e t h e a l p h a b e t s a n d n u m b er er s . 4.Print the preprocessor, function, keyword using yytext.lex tool. 5.Print the relational, assignment and all the operator using yytext() tool. 6 . A l so s o s c a n a n d p r i n t w h e r e t h e l o o p e n d s a n d b e g i n s. s. 7.Use yywrap() to enter an error. 8.Stop the program. Program: %{ int COMMENT=0; %} identifier [_a-zA-Z][_a-zA-Z0-9]* %% #.* {printf("\n%s is a PREPROCESSOR DIRECTIVE",yytext);} int | float | char | while | for | do | if | break |




continue | void | switch | case | long | struct | const | typedef | return | else | goto {printf("\n\t%s is a KEYWORD",yytext);} "/*" {COMMENT=1; printf("\n\n\t COMMENT ENDS\n");} {identifier}$ {if(!COMMENT) printf("\n\nFUNCTION \n\t%s",yytext);} \n\t%s",yytext) ;} \{ {if(!COMMENT) printf("\n BLOCK BEGINS");} \} {if(!COMMENT) printf("\n BLOCK ENDS");} {identifier}(\[[0-9]*\])? {if(!COMMENT) printf("\n %s IDENTIFIER",yytext);} \".*\" {if(!COMMENT) printf("\n %s is a STRING",yytext);} [0-9]+ {if(!COMMENT) printf("\n %s is a NUMBER",yytext);} $(\;)? {if(!COMMENT) PRINTF("\N\T");ECHO; printf("\n");} \( ECHO; = {if(!COMMENT) printf("\n\t%s is an ASSIGNMENT OPERATOR",yytext);} \<= | \>= | \< | == | \> {if(!COMMENT) printf("\n\t%s is a RELATIONAL OPERATOR",yytext);} |\n %% int main(int argc,char **argv) { if(argc > 1) { FILE *file; file=fopen(argv[1],"r"); if(!file) { printf("could not open o pen %s\n",argv[1]); exit(0); } yyin=file; } yylex(); printf("\n\n"); return 0; } int yywrap()



{ return 0; } INPUT(area.c): # include #include double area_of_circle(double r); int main(int arge.char*argv[]) { if(arge<2) { printf(“usage:%s printf(“usage:%s radius\n”,argv[0]); radius\n”,argv[0]); { exit(1); } else { double radius=atof(argv[1]); double area=area_of_circle(radius); printf("Area of circle with radius % f=%f\n",radius,area); } return 0; } OUTPUT








Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 4A 4A YACC YA CC PROG PROGRA RAM M TO RECOGNI RECOGNIZE ZE A VALID VALID ARITHME ARITHMETIC TICEX EXPR PRES ESSI SION ON Aim:

To write a YACC program to recognize a valid arithmetic expression that uses operators +, -, * and /. Steps: 1.Start the program 2. Open a file filename.c in read and include the yylex() tool for input scanning. 3. Define the arithmetic expression 4. Also scan and print where the loop ends and begins. 5 . U s e y a c c t o p r i n t t h e va valid arit rithmetic expres expressi sion. on. 6. Stop the program. Program: LEX %{ #include"y.tab.h"extern yylval; %} %% [0-9]+ {yylval=atoi(yytext); return NUMBER;} [a-zA-Z]+ {return ID;} [\t]+ ; \n {return 0;} . {return yytext[0];} %% YACC %{ #include %} %token NUMBER ID %left '+' '-' %left '*' '/' %% expr: expr '+' expr |expr '-' expr |expr '*' expr |expr '/' expr |'-'NUMBER| '-'ID |'('expr')' |NUMBER |ID;



%% main() { printf("Enter the expression\n"); expressio n\n"); yyparse(); printf("\nExpression is valid\n"); exit(0); } int yyerror(char *s) { printf("\nExpression is invalid"); exit(0); } Criteria for evaluation:

$lex p4a.l $yacc – $yacc –d d p4a.y $cc lex.yy.c y.tab.c –ll –ll $./a.out Enter Enter the express expression(a*b ion(a*b+5 +5)) Expression is valid $./a.out Enter the expression (a+6-) Expression is invalid





Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 4B 4B YACC PROGRAM TO RECOGNIZE A VALID VARIABLE Aim:

To write a YACC program to recognize a valid variable, which starts with a letter, followed by any number of letters or digits. Steps: 1 . Start the program 2. Open a file filename.c in read and include the yylex() tool for input scanning. 3. Define the string variable 4. Also scan and print where the loop ends and begins. 5 . U s e y a c c t o p r i n t t h e string is is valid or not. 6. Stop the program. Program:

LEX %{ #include"y.tab.h"extern yylval; %} %% [0-9]+ {yylval=atoi(yytext); return DIGIT;} [a-zA-Z]+ {return LETTER;} [\t] ; \n return 0; .{return yytext[0];} %% YACC %{ #include %} %token LETTER DIGIT %% variable: LETTER|LETTER rest ; rest: LETTER rest |DIGIT rest |LETTER |DIGIT



; %% main() { yyparse(); printf("The string is a valid variable\n"); variable\n") ; } int yyerror(char *s) { printf("this is not a valid variable\n"); exit(0); } Criteria for evaluation: $lex p4b.l $yacc – $yacc – d p4b.y $cc lex.yy.c y.tab.c – y.tab.c – ll ll $./a.out input34 The string is a valid variable $./a.out 89file This is not a valid variable.





Result: Thus the above the program is executed and the required output is obtained.

EXERCISE – 4C 4C IMPLEMENTATION OF CALCULATION USING LEX ANDYACC Aim: To write a program to implement calculator using lex and yacc. Steps: 1.Start the program. 2.Perform the calculation using both the lex and yacc. 3.In the lex tool, if the given expression contains numbers and letters then they are displayed. 4.In the same way, the digits, letters and uminus are identified and displayed using yacctool. 5.The calculation is performed and the result is displayed. 6.Stop the program. Program: USING LEX TOOL: %{ #include #include"y.tab.h" int c; extern int yylaval; %} %% ""; [a-z] { c=yytext[0]; yylaval=c-'a'; return(LETTER); } [0-9] { c=yytext[0]; yylaval=c-'0'; return(DIGIT);


} [a-z0-9/b] { c=yytext[0]; return(c); } USING YACC TOOL: %{ #include int regs[26]; int base; %} %start list %token DIGIT LETTER %left'|' %left'&' %left'+''-' %left'*''/''%' %left UMINUS %% list: | list stat'\n' |list error'\n' { yyerror(); }; stat:expr { printf("%d\n",$1); } | LETTER'='expr { regs[$1]=$3; }; expr:'('expr')' { $$=$2; }

| expr'*'expr { $$=$1*$3; } |




expr'/'expr { $$=$1/$3; } | expr'%'expr { $$=$1%$3; } | expr'+'expr { $$=$1+$3; } | expr'-'expr { $$=$1-$3; } | expr'&'expr { $$=$1&$3; } | expr'|'expr { $$=$1|$3; } | '-'expr%prec UMINUS { $$=-$2; } | LETTER { $$=regs[$1]; } | number ; number:DIGIT { $$=$1; base=($1==0)?8:10; base=($1==0)?8:10 ; } |




number DIGIT { $$=base*$1+$2; } %% main() { return(yyparse()); } yyerror(s) char*s; { fprintf(stderr,"%s\n",s); } yywrap() { return(1); } OUTPUT






Result: Thus the above the program is executed and the required output is obtained EXERCISE – 5 ABSTRACT SYNTAX TREE Aim: Convert the BNF rules into Yacc form and write code to generate Abstract Syntax Tree. Steps: 1. Start the program. 2. Include the yylex() yylex() tool for open a file file as & & . 3 . A l s o s c a n a n d p r i n t w h e r e t h e lo lo o p e n d s a n d b e g i n s . 4 . U s e y a c c t o p r i n t t h e a b s t r a c t s y nt nt a x t r e e 5 . Stop Stop the program. Program: %{ #include"y.tab.h" #include #include int LineNo=1; %}




identifier [a-zA-Z][_a-zA-Z0-9]* number [0-9]+|([0-9]*\.[0-9]+) %% main return MAIN; if return IF; else return ELSE; while return WHILE; int | char | float return TYPE; {identifier} {strcpy(yylval.var,yytext); return VAR;} {number} {strcpy(yylval.var,yytext); return NUM;} \< | \> | \>= | \<= | == {strcpy(yylval.var,yytext); return RELOP;} [ \t] ; \n LineNo++; . return yytext[0]; %% %{ #include #include struct quad{ char op[5]; char arg1[10]; char arg2[10]; char result[10]; }QUAD[30]; struct stack { int items[100]; int top; }stk; int Index=0,tIndex=0,StNo,Ind,tInd; extern int LineNo; %} %union { char var[10]; } %token NUM VAR RELOP %token MAIN IF ELSE WHILE TYPE %type EXPR ASSIGNMENT CONDITION IFST ELSEST WHILELOOP WHILELOOP Cseitquestions.blogspot.in Cseitquestions.blogspot.in %left '-' '+' %left '*' '/' %% PROGRAM : MAIN BLOCK ; BLOCK: '{' CODE '}' ; CODE: BLOCK | STATEMENT CODE | STATEMENT; STATEMENT: DESCT ';' | ASSIGNMENT ';‟ | CONDST | WHILEST ; DESCT: TYPE VARLIST ; VARLIST: VAR ',' VARLIST | VAR ; ASSIGNMENT: VAR '=' EXPR{ strcpy(QUAD[Index].op,"="); strcpy(QUAD[Index].arg1,$3); strcpy(QUAD[Index].arg2,""); strcpy(QUAD[Index].result,$1); strcpy($$,QUAD[Index++].result); } ; EXPR: EXPR '+' EXPR {AddQuadruple("+",$1,$3,$$);} |EXPR '-' EXPR {AddQuadruple("-",$1,$3,$$);} | EXPR '*' EXPR {AddQuadruple("*",$1,$3,$$);} | EXPR '/' EXPR {AddQuadruple("/",$1,$3,$$);} | '-' EXPR {AddQuadruple("UMIN",$2,"",$$);} | '(' EXPR ')' {strcpy($$,$2);} | VAR | NUM ; CONDST:IFST{ Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index); Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index);} | IFST ELSEST ; IFST: IF '(' CONDITION ')'{ strcpy(QUAD[Index].op,"=="); strcpy(QUAD[Index].arg1,$3); Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in strcpy(QUAD[Index].arg2,"FALSE"); strcpy(QUAD[Index].result,"-1"); push(Index); Index++; } BLOCK{ strcpy(QUAD[Index].op,"GOTO"); strcpy(QUAD[Index].arg1,""); strcpy(QUAD[Index].arg2,""); strcpy(QUAD[Index].result,"-1"); push(Index); Index++ ; } ;ELSEST: ELSE{ tInd=pop(); Ind=pop(); push(tInd); sprintf(QUAD[Ind].result,"%d",Index); } BLOCK{ Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index); } ; CONDITION: VAR RELOP VAR {AddQuadruple($2,$1,$3,$$); StNo=Index-1; } | VAR | NUM ; WHILEST:WHILELOOP{ Ind=pop(); sprintf(QUAD[Ind].result,"%d",StNo); Ind=pop(); sprintf(QUAD[Ind].result,"%d",Index); } ; WHILELOOP: WHILE '(' CONDITION ')' { strcpy(QUAD[Index].op,"=="); strcpy(QUAD[Index].arg1,$3); strcpy(QUAD[Index].arg2,"FALSE"); strcpy(QUAD[Index].result,"-1"); push(Index); Index++; } BLOCK { strcpy(QUAD[Index].op,"GOTO"); Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in strcpy(QUAD[Index].arg1,""); strcpy(QUAD[Index].arg2,""); strcpy(QUAD[Index].result,"-1"); push(Index); Index++; }; %% extern FILE *yyin; int main(int argc,char *argv[]) { FILE *fp; int i; if(argc>1) { fp=fopen(argv[1],"r"); if(!fp) { printf("\n File not found"); found") ; exit(0); } yyin=fp; } yyparse(); printf("\n\n\t\t ----------------------------""\n\t\t ---------------------- ------""\n\t\t Pos Operator Result" "\n\t\t--------------------"); for(i=0;i Arg1 Arg2 Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in { printf("\n Stack underflow\n"); u nderflow\n"); exit(0); } data=stk.items[stk.top--]; return data; } void AddQuadruple(char op[5],char arg1[10],char arg2[10],char result[10]) { strcpy(QUAD[Index].op,op); strcpy(QUAD[Index].arg1,arg1); strcpy(QUAD[Index].arg2,arg2); sprintf(QUAD[Index].result,"t%d",tIndex++); strcpy(result,QUAD[Index++].result); } yyerror() { printf("\n Error on line no:%d",LineNo); no:%d ",LineNo); } Input: $vi test.c main() { int a,b,c; if(a Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 6 IMPLEMENT TYPE CHECKING Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Aim: To check whether string belongs to a grammar or not Steps: 1 Start the program. 2 Declare two character arrays str[],token[] and initialize integer variables a=0,b=0,c,d. 3. Input the string from the user. 4. Print the token generated for the string . 5. Compare tokens and store the result in d. 6. If d=0 then print that the string is in the grammar. 7. Else print that the string is not in the grammar. 8. Stop the program. Program: #include #include #include #include void main() { int a=0,b=0,c; char str[20],tok[11]; clrscr(); printf("Input the expression = "); gets(str); while(str[a]!='\0') { if((str[a]=='(')||(str[a]=='{')) { tok[b]='4'; b++; } if((str[a]==')')||(str[a]=='}')) { tok[b]='5'; b++; } if(isdigit(str[a])) { while(isdigit(str[a])) { a++; } a--; tok[b]='6'; b++; } if(str[a]=='+') { Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in tok[b]='2'; b++; } if(str[a]=='*') { tok[b]='3'; b++; } a++; } tok[b]='\0'; puts(tok); puts(tok) ; b=0; while(tok[b]!='\0') { if(((tok[b]=='6')&&(tok[b+1]=='2')&&(tok[b+2]=='6'))||((tok[b]=='6')&&(tok[b+1 ]=='3')&&(tok[b+2]=='6'))||((tok[b]=='4')&&(tok[b+1]=='6')&&(tok[b+2]=='5'))/*||((tok[b ]!=6)&&(tok[b+1]!='\0'))*/) { tok[b]='6'; c=b+1; while(tok[c]!='\0') { tok[c]=tok[c+2]; c++; } tok[c]='\0'; puts(tok); puts(tok) ; b=0; } else { b++; puts(tok); puts(tok) ; } } int d; d=strcmp(tok,"6"); if(d==0) { printf("It is in the gra mmar."); } else {printf("It is not in the grammar."); } getch(); } Cseitquestions.blogspot.in Cseitquestions.blogspot.in Criteria for evaluation: Input the expression = (23+) 4625 4625 4625 4625 4625 It is not in the grammar. Input the expression = (2+(3+4)+5) 46246265265 46246265265 46246265265 46246265265 46246265265 462465265 462465265 462465265 462465265 4626265 4626265 46265 46265 465 6 6 It is in the grammar. Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 7 IMPLEMENTATION OF STACK STORAGE ALLOCATION Aim: To write a program to implement a stack storage using array. Steps: INSERTION PUSH(item) 1. If (item = max of stack) Print “overflow” Return 2. top = top + 1 3. stack[top] = item 4. Return DELETION POP(item) 1. If (top = - 1) Print “underflow” Return 2. Item = stack[top] 3. top = top – top – 1 1 4. Return DISPLAY 1. If top = - 1 Print “underflow” 2. repeat step 3 for i = top to i >= 0 3.Print 3. Print stack[i] 4. Return Program: #include #include #define MAXSIZE10 void push(); int pop(); void traverse(); int stack[MAXSIZE]; int Top=-1; void main() { int choice; char ch; do Cseitquestions.blogspot.in Cseitquestions.blogspot.in { clrscr(); printf("\n1.PUSH"); printf("\n2. POP "); printf("\n3. TRAVERSE "); printf("\nEnter printf("\ nEnter your choice"); scanf("%d",&choice); switch(choice) { case 1: push(); break; case 2: printf("\nThe deleted element is %d",pop()); break; case 3: traverse(); break; default: printf("\nYou Entered Wrong Choice"); } printf("\nDo You Wish To Continue(Y/N)"); Co ntinue(Y/N)"); fflush(stdin); scanf("%c",&ch); } while(ch=='Y' || ch=='y'); } void push() { int item; if(Top == MAXSIZE - 1) { printf("\nThe Stack IsFull"); getch(); exit(0); } else { printf("Enter the element e lement to beinserted"); scanf("%d",&item); Top= Top+1; stack[Top] = item; } } int pop() { int item; if(Top == -1) { printf("The stack is Empty"); Empt y"); getch(); exit(0); } else { item = stack[Top]; Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Top = Top-1; } return(item); } void traverse() { int i; if(Top == -1) { printf("The Stack is Empty"); getch(); exit(0); } else { for(i=Top;i>=0;i--) { printf("Traverse the element"); e lement"); printf("\n%d",stack[i]); } } } OUTPUT Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 8 CONSTRUCTION CONSTRUCTION OF DAG Aim: To write a program to construct Direct Acyclic Graph using c. Steps: 1. Start the program. 2. Include the structure for node creation & construct the tree. 3 . A l s o s c a n a n d p r i n t w h e r e t h e lo lo o p e n d s a n d b e g i n s . 4 . p r i nt t he Direc Directt Acycl cycliic Graph Graph . 5 . Stop the program. Program: #include #include struct node { struct node *next; char id[10]; char left[10]; char right[10]; char attach[3][10]; }; struct node *head; file *f; int i,s=0; char str[25],store[10][25]; void construct_tree() { struct node *temp; struct node *t; struct node *ptr; int flag=0,fl=0; temp=head; if(s==5 || s==6) { Cseitquestions.blogspot.in Cseitquestions.blogspot.in while(temp->next!=null) { if(!strcmp(store[2],temp->next->left)) flag+=1; if(!strcmp(store[4],temp->next->right)) flag+=2; if(flag!=0) break; temp=temp->next; } t=head; while(t->next!=null) t=t->next; if(flag==0) { ptr=(struct node*)malloc(sizeof(struct node*)malloc(sizeo f(struct node)); nod e)); t->next=ptr; if(s==5) strcpy(ptr->id,store[3]); else strcpy(ptr->id,strcat(store[3],store[5])); t=head; while(t->next!=null) { if(!strcmp(t->next->attach[0],store[2])) { f1=1; break; } if(strcmp(t->next->attach[1],””)) if(strcmp(t->next->attach[1],””)) if(!strcmp(t->next->attach[1],store[2])) { f1 = 1; break; } t=t->next; } if(f1) strcpy(ptr->left,t->next->id); else strcpy(ptr->left,store[2]); f1=0; t=head; while(t->next!=null) { if(!strcmp(t->next->attach[0],store[4])) { Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in f1 = 1; break; } if(strcmp(t->next->attach[1],””)) if(strcmp(t->next->attach[1],””)) if(!strcmp(t->next->attach[1],store[4])) { f1= 1; break; } t=t->next; } if(f1) strcpy(ptr->right,t->next->id); else strcpy(ptr->right,store[0]); strcpy(ptr->attach[1],,””); strcpy(ptr->attach[1],,””); ptr->next=null; } else if(flag==3) strcpy(temp->next->attack[1],store[0]; } if(s==3) { while(temp->next!=null) { if(!strcmp(store[2],temp->next->attach[0])) break; temp=temp->next; } strcpy(temp->next->attach[1],store[0]); } } int main() { struct node *temp; struct node *t; clrscr(); f=fopen(“input.text”,”r”); head=(struct node*)malloc(size of (struct node)); head->next=null; while(!feof(f)) { fscanf(f,”%s”,str); if(!strcmp(str,”;”)) { construct _tree(); s=0; Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in }else strcpy(store[s++],str); }printf(“\n\nid\tleft\tright\tattached }printf(“\n\nid\tleft\tright\tattached ids\n\n”); ids\n\n”); temp=head; while(temp->next!=null) { printf(“\n\n%s\t%s\t%s\ printf(“\n\n%s\t%s\t%s\t%s,”temp t%s,”temp->next->id, ->next->id, temp->next->left,temp->next->right,temp->next->attach[0]); if(strcmp(temp->next->attach[1],””)) if(strcmp(temp->next->attach[1],””)) printf(“\ printf(“\t%s”,temp->next->attach[1]); t%s”,temp->next->attach[1]); temp=temp->next; } getch(); return 0;} Criteria for evaluation: INPUT: t1 = 4 * I; t2 = a[t1]; t3 = 4 *I; t4= b [t3]; t5= t2 *t4; t6 = prod + t5; prod = 16; t7 = i+1; i=t7; OUTPUT: ID LEFT * 4 [] a [] b * [] + prod + I RIGHT I * * [] * | ATTACHED t1 t2 t4 t5 t6 t7 IDs t3 prod i Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 9 IMPLEMENTATION THE BACK END OF COMPILER COMPILER Aim: To write a C program to implement the Back end of the compiler. Steps: 1)Start the program. 2)Get the three variables from statements and stored in the text file k.txt. 3)Compile the program and give the path of the source file. 4)Execute the program. 5)Target code for the given statement was produced. 6)Stop the program. Program: #include #include #include #include void main() { int i=2,j=0,k=2,k1=0; char ip[10],kk[10]; FILE *fp;clrscr(); printf("\nEnter t he filename of the intermediate code"); scanf("%s",&kk); fp=fopen(kk,"r"); if(fp==NULL) { printf("\nError in Opening the file"); Cseitquestions.blogspot.in Cseitquestions.blogspot.in getch(); }clrscr(); while(!feof(fp)) { fscanf(fp,"%s\n",ip); printf("\t\t%s\n",ip); } rewind(fp); printf("\n----------------------------- printf("\n---------- -------------------- \n"); printf("\tStatement printf("\tStat ement \t\t target code\n"); printf("\n----------------------------- printf("\n----- ------------------------- \n"); while(!feof(fp)) { fscanf(fp,"%s",ip); printf("\t%s",ip); printf("\t\tMOV printf("\t\t MOV %c,R%d\n\t",ip[i+k],j); if(ip[i+1]=='+') printf("\t\tADD"); else printf("\t\tSUB"); if(islower(ip[i])) printf("%c,R%d\n\n",ip[i+k1],j); printf("%c,R%d\n\n",ip[i+k1] ,j); else printf("%c,%c\n",ip[i],ip[i+2]); printf("%c,%c\n",ip[i],ip[ i+2]); j++; k1=2; k=0;} printf("\n------------------------------ printf("\n---------- ---------------------\n"); \n"); getch(); fclose(fp); } OUTPUT Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 10 10 IMPLEMENTATION THE FRONT END OF COMPILER Aim: To write a C program to implement the front end of the compiler. Steps: 1) Start the program. 2) Get the coding from the u ser. 3) Find the operators, arguments and results from the coding. Cseitquestions.blogspot.in Cseitquestions.blogspot.in 4) Display the value in the table. 5) Stop the program. Program: #include #include #include void main() { char pg[100][100],str1[24]; int tem=-1,ct=0,i=-1,j=0,j1,pos=-1,t=-1,flag,flag1,tt=0,fg=0; clrscr(); printf("Enter the codings co dings \n"); while(i>-2) { i++; lab1: t++; scanf("%s",&pg[i]); if((strcmp(pg[i],"getch();"))==0) { i=-2; goto lab1; } } printf("\n pos \t o per \t arg1 \t arg2 \tresult \n"); \ n"); while(j Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in if(pg[j][3]=='=') printf("%d\t%c\t%c\t%c\tt%%d\n",pos,pg[j][2],pg[ printf("%d\t%c\t %c\t%c\tt%%d\n",pos,pg[j][2],pg[j][1],pg[j][4],tem); j][1],pg[j][4],tem); else printf("%d\t%c\t%c\t%c\tt%d\n",pos,pg[j][2],pg[ printf("%d\t%c\t %c\t%c\tt%d\n",pos,pg[j][2],pg[j][1],pg[j][3],tem); j][1],pg[j][3],tem); j1+=2; pos++; while((strcmp(pg[j],"}"))!=0) { j++; if(pg[j][1]=='=') { tt=j; fg=1; } ct++; } ct=ct+pos+1; printf("%d\t==\tt%d\tFALSE\t%d\n",pos,tem,ct); printf("%d\t==\tt%d \tFALSE\t%d\n",pos,tem,ct); if(fg==1) { j=tt; goto lop; } while((strcmp(pg[j],"}"))!=0) { pos++; tem++; printf("%d\t%c\t%c\t%c\tt%d\n",pos,pg[j][3],pg[ printf("%d\t%c\t %c\t%c\tt%d\n",pos,pg[j][3],pg[j][2],pg[j][4],tem); j][2],pg[j][4],tem); pos++; printf("%d\t:=\tt%d\t\t%c\n",pos,tem,pg[j][0]); printf("%d\t:=\tt %d\t\t%c\n",pos,tem,pg[j][0]); j++; } if((strcmp(pg[j+1],"else"))==0) { ct=0; j++; j1=j; j1+=2; pos++; while((strcmp(pg[j],"}"))!=0) { j1++; ct++; } ct=ct*2; ct++; ct+=(pos+1); j+=2; Cseitquestions.blogspot.in Cseitquestions.blogspot.in printf("%d\tGOTO\t\t\\t%d\n",pos,ct); printf("%d\tGOTO\t\t\ \t%d\n",pos,ct); while((strcmp(pg[j],"}"))!=0) { pos++; tem++; printf("%d\t%c\t%c\t%c\tt%d\n",pos,pg[j][3],pg[ printf("%d\t%c\t %c\t%c\tt%d\n",pos,pg[j][3],pg[j][2],pg[j][4],tem); j][2],pg[j][4],tem); pos++; printf("%t:=\tt%d\t\t%c\n",pos,tem,pg[j][0]); printf("%t:=\tt%d\t\t %c\n",pos,tem,pg[j][0]); j++; } pos++; printf("%d\tGOTO\t\t\t\%d\n",pos,ct); printf("%d\tGOTO\t\t\t \%d\n",pos,ct); } } if((strcmp(pg[j],"}"))==0) { pos++; printf("%d\tGOTO\t\t\t%d\n",pos,pos+1); printf("%d\tGOTO\t\t\t%d\ n",pos,pos+1); } j++; } getch(); } OUTPUT Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the above the program is executed and the required output is obtained. EXERCISE – 11 11 IMPLEMENTATION IMPLEMENTATION OF SIMPLE CODE OPTIMIZATION TECHNIQUES Aim: To write a program for implementation of Code Optimization Technique in for and do-while loop using C++. Steps: 1.Generate the program for factorial program using for and do-while loop to specify optimization technique. 2.In for loop variable initialization is activated first and the condition is checked next. If the condition is true the corresponding statements are executed and specified increment / decrement operation is performed. 3.The for loop operation is activated till the condition failure. 4.In do-while loop the variable is initialized and the statements are executed then the condition checking and increment / decrement operation is performed. 5.When comparing both for and do-while loop for optimization do-while is best because first the statement execution is done then only the condition is checked. So, during the statement execution itself we can fin the inconvenience of the result and no need to wait for the specified condition result. 6.Finally when considering Code Optimization in loop do-while is best with respect to performance. Program: Before: Using for: #include #include int main() { int i, n; int fact=1; cout<<"\nEnter a number: "; cin>>n; for(i=n ; i>=1 ; i--) fact = fact * i; Cseitquestions.blogspot.in Cseitquestions.blogspot.in cout<<"The factoral value is: "< #include void main() { clrscr(); int n,f; f=1; cout<<"Enter the number:\n"; cin>>n; do { f=f*n; n--; }while(n>0); cout<<"The factorial value is:"< Criteria for evaluation: USING for LOOP Enter the number: 5 The factorial value is: 120 USING do-while LOOP Enter the number: 3 The factorial value is: 6 Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Cseitquestions.blogspot.in Result: Thus the program for implementation of Code Optimization technique is executed and verified.

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