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Saturday, 9 March 2019

MEC306 : DESIGN OF MACHINE ELEMENTS -2 Study Material Notes - MEC 306


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March 09, 2019

Saturday, 23 February 2019

Program to count the number of vowels and consonants in a given string | Compiler Design | Lex Programs

Compiler Design Lap Program :  %{  #include<stdio.h>  int vowels=0;  int cons=0; %} %% [aeiouAEIOU] {vowels++;} ...

Compiler Design Lap Program : 

%{
 #include<stdio.h>
 int vowels=0;
 int cons=0;
%}
%%
[aeiouAEIOU] {vowels++;}
[a-zA-Z] {cons++;}
%%
int yywrap()
{
return 1;
}
main()
{
 printf(“Enter the string.. at end press ^d\n”);
 yylex();
 printf(“No of vowels=%d\nNo of consonants=%d\n”,vowels,cons);
} 
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February 23, 2019

Program to count the number of characters, words, spaces, end of lines in a given input file. | Compiler Design Programs | Lex Programs

Compiler Design Lab Program :  %{  #include<stdio.h>  Int c=0, w=0, s=0, l=0; %} WORD [^ \t\n,\.:]+ EOL [\n] BLA...

Compiler Design Lab Program : 

%{
 #include<stdio.h>
 Int c=0, w=0, s=0, l=0;
%}
WORD [^ \t\n,\.:]+
EOL [\n]
BLANK [ ]
%%
{WORD} {w++; c=c+yyleng;}
{BLANK} {s++;}
{EOL} {l++;}
. {c++;}
%%
int yywrap()
{
return 1;
}
main(int argc, char *argv[])
{
 If(argc!=2)
 {
 printf(“Usage: <./a.out> <sourcefile>\n”);
 exit(0);
 }
 yyin=fopen(argv[1],”r”);
yylex();
printf(“No of characters=%d\nNo of words=%d\nNo of
spaces=%d\n No of lines=%d”,c,w,s,l);
} 
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February 23, 2019

Program to count no of: a) +ve and –ve integers b) +ve and –ve fractions | Compiler Design Programs | Lex Programs

Compiler Design Lap Program : %{  #include<stdio.h>  int posint=0, negint=0,posfraction=0, negfraction=0; %} %% [-...

Compiler Design Lap Program :

%{
 #include<stdio.h>
 int posint=0, negint=0,posfraction=0, negfraction=0;
%}
%%
[-][0-9]+ {negint++;}
[+]?[0-9]+ {posint++;}
[+]?[0-9]*\.[0-9]+ {posfraction++;}
[-][0-9]* \.[0-9]+ {negfraction++;}
%%
int yywrap()
{
return 1;
}
main(int argc, char *argv[])
{
 If(argc!=2)
 {
 printf(“Usage: <./a.out> <sourcefile>\n”);
 exit(0);
 }
 yyin=fopen(argv[1],”r”);
yylex();
printf(“No of +ve integers=%d\n No of –ve integers=%d\n No of
+ve
fractions=%d\n No of –ve fractions=%d\n”, posint, negint,
posfraction, negfraction);
}
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February 23, 2019

Program to count the no of comment line in a given C program. Also eliminate them and copy that program into separate file . | Compiler Design Lab Programs | Lex Programs

Compiler Design Lap Program : %{  #include<stdio.h>  int com=0; %} %s COMMENT %% “/*”[.]*”*/” {com++;} “/*” {B...

Compiler Design Lap Program :

%{
 #include<stdio.h>
 int com=0;
%}
%s COMMENT
%%
“/*”[.]*”*/” {com++;}
“/*” {BEGIN COMMENT ;}
<COMMENT>”*/” {BEGIN 0; com++ ;}
<COMMENT>\n {com++ ;}
<COMMENT>. {;}
.|\n {fprintf(yyout,”%s”,yytext);
%%
int yywrap()
{
return 1;
}
main(int argc, char *argv[])
{
 If(argc!=2)
 {
 printf(“Usage: <./a.out> <sourcefile> <destn file>\n”);
 exit(0);
 }
 yyin=fopen(argv[1],”r”);
 yyout=fopen(argv[2],”w”);
 yylex();
 printf(“No of comment lines=%d\n”,com);
} 
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February 23, 2019

Program to count the no of ‘scanf’ and ‘printf’ statements in a C program. Replace them with ‘readf’ and ‘writef’ statements respectively. | Compiler Design Programs | Lex Programs

Compiler Design Program Lab : %{ #include<stdio.h> int pc=0, sc=0; %} %% “printf” { fprintf(yyout,”writef”); pc++;...

Compiler Design Program Lab :

%{
#include<stdio.h>
int pc=0, sc=0;
%}
%%
“printf” { fprintf(yyout,”writef”); pc++;}
“scanf” { fprintf(yyout,”readf”); sc++;}
%%
int yywrap()
{
return 1;
}
main(int argc, char *argv[])
{
 if(argc!=2)
 {
 printf(“Usage: <./a.out> <sourcefile> <destn file>\n”);
 exit(0);
 }
 yyin=fopen(argv[1],”r”);
 yyout=fopen(argv[2],”w”);
 yylex();
 printf(“No of printf statements = %d\n No of scanf
statements=%d\n”, pc, sc);
} 
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February 23, 2019

Program to recognize a valid arithmetic expression and identify the identifiers and operators present. Print them separately. | Compiler Design Lab Programs | Lex Programs

Compiler Design Program : %{  #include<stdio.h>  #include<string.h>  int noprt=0, nopnd=0, valid=1, top=-1, m, l...

Compiler Design Program :

%{
 #include<stdio.h>
 #include<string.h>
 int noprt=0, nopnd=0, valid=1, top=-1, m, l=0, j=0;
 char opnd[10][10], oprt[10][10], a[100];
%}
%%
“(“ { top++; a[top]=’(‘ ; }
“{“ { top++; a[top]=’{‘ ; }
“[“ { top++; a[top]=’[‘ ; }
“)” { if(a[top]!=’(‘)
 {
 valid=0; return;
}
else
 top--;
 }
“}” { if(a[top]!=’{‘)
 {
 valid=0; return;
}
else
 top--;
 }
“]” { if(a[top]!=’[‘)
 {
 valid=0; return;
}
else
 top--;
 }
“+”|”-“|”*”|”/” { noprt++;
 strcpy(oprt[l], yytext);
l++;
 }
[0-9]+|[a-zA-Z][a-zA-Z0-9_]* {nopnd++; 
strcpy(opnd[j],yytext);
 j++;
 }
%%
int yywrap()
{
return 1;
}
main()
{
 int k;
 printf(“Enter the expression.. at end press ^d\n”);
 yylex();
 if(valid==1 && i==-1 && (nopnd-noprt)==1)
 {
 printf(“The expression is valid\n”);
 printf(“The operators are\n”);
 for(k=0;k<l;k++)
 Printf(“%s\n”,oprt[k]);
 for(k=0;k<l;k++)
 Printf(“%s\n”,opnd[k]);
 }
 else
 Printf(“The expression is invalid”);
}
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February 23, 2019

 

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