Operator precedence parsing algorithm

//This is operator precedence parsing algorithm using C

char pre[10][10],ss[30],rs[30],stk[10][2];

int i,j,k,n,p,sp=0,br=0;

void push();

void pop();

char chk_pre();

void reduce_ss();

void main()

{

clrscr();

strcpy(pre[0]," +*()]");

strcpy(pre[1],"+><<>>");

strcpy(pre[2],"*>><>>");

strcpy(pre[3],"(<<<=e");

strcpy(pre[4],")>>e>>");

strcpy(pre[5],"[<<<e=");

n=6;

scanf ("%s",ss, printf("Enter the source string : "));

strcpy(rs,"[");

strcat(rs,ss);

strcat(rs,"]");

strcpy(ss,rs);

printf("\n");

for(i=0;i<n;i++)

{

for(j=0;j<strlen(pre[i]);j++)

printf("%c\t",pre[i][j]);

printf("\n");}

while(ss[2]!=']' && br==0)

{

sp=0;

for(i=0;i<strlen(ss);i++)

{

if( (ss[i]>='a' && ss[i]<='z') || (ss[i]>='A' && ss[i]<='Z') )

continue;

push();

if(stk[sp-1][1]=='e' || ss[1]==']' || ss[0]==']')

{ printf("\n\nInvalid string.");

br=1;

break; }

if((stk[sp-1][1]=='>' && stk[sp-2][1]=='<') || stk[sp-1][1]=='=')

{ reduce_ss();

pop(); } } }

if(ss[2]==']')

printf("\nValid String.");

getch();

}

void push()

{ stk[sp][0]=ss[i];

stk[sp][1]=chk_pre();

sp++; }

void pop()

{ char tmp,tmp1;

sp--;

tmp=stk[sp][0];

tmp1=stk[sp][1];

sp--;

stk[sp][0]=tmp;

stk[sp][1]=tmp1;

sp++; }

char chk_pre()

{ char tmp;

if(i==0)

tmp=' ';

else

{ for(j=0;j<n;j++)

for(k=0;k<strlen(pre[j]);k++)

if(stk[sp-1][0]==pre[j][0] && stk[sp][0]==pre[0][k])

tmp=(pre[j][k]); }

return tmp; }

void reduce_ss()

{  char t1[30],t3[30];

int tmp;

if(stk[sp-1][0]==')' && stk[sp-2][0]=='(')

tmp=i-2;

else

tmp=i-3;

for(j=0;j<tmp;j++)

{  t1[j]=ss[j];  }

t1[j]='\0';

tmp=tmp+3;

k=0;

for(j=tmp;j<strlen(ss);j++)

{  t3[k]=ss[j];

k++;  }

t3[k]='\0';

strcpy(ss,"");

strcpy(ss,t1);

strcat(ss,"i");

strcat(ss,t3);

}

Database connection in classic ASP

Dim con
Dim rs
Dim ConnectString
dim accessdb
set con=server.CreateObject("ADODB.Connection")
set rs=server.CreateObject("ADODB.RecordSet")
ConnectString = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=C:\Inetpub\wwwroot\database.mdb"
con.open ConnectString
accessdb="database.mdb"
mysql="select * from table"

rs.open mysql, con, 3, 3

Advance microprocessor 8086 programs

Arithmetic operation

Practical-1

Aim: Write a program to addition of two numbers.

Software requirements: TASM Simulator

Logic:

assume cs:code,ds:data

data segment

a dw 0001h

b dw 0005h

sum dw ?

data ends

code segment

start: mov ax,data

mov ds,ax

mov ax,a

mov bx,b

add ax,bx

mov sum,ax

mov ah,04ch

int 21h

code ends

end start

Output:

Conclusion:

The program of  addition of two numbers is successfully executed with the use of tasam simulator.

Practical-2

Aim: Write a program to multiply two 16 bit numbers.

Software requirements: TASM Simulator

Logic:

assume cs:code,ds:data

data segment

a dw 0002h

b dw 0005h

multi dw ?

data ends

code segment

start:mov ax,data

mov ds,ax

mov ax,a

mov bx,b

mul bx

mov multi,ax

mov ah,04ch

int 21h

code ends

end start

Output:

Conclusion:

The program of to multiply two 16 bit numbers is successfully executed with the use of tasam simulator.

Practical-3

Aim: Write a program to add and subtract two 4 digit BCD numbers.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

A DB 41H

B DB 29H

SUM DB ?

DIF DB ?

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV AL,A

MOV CL,AL

MOV BL,B

ADD AL,BL

DAA

MOV SUM,AL

MOv AL,CL

SUB AL,BL

DAS

MOV DIF,AL

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of  add and subtract two 4 digit BCD numbers is successfully executed with the use of tasam simulator

Practical-4

Aim: Write a program to find sum of an integer 8-bit array.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

A DB 01H,05H,06H

SUM DB ?

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV CL,03H

MOV AL,00H

MOV BL,00H

LEA SI,A

AGAIN:

MOV BL,[SI]

ADD AL,BL

INC SI

LOOP AGAIN

DAA

MOV SUM,AL

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of  find sum of an integer 8-bit array is successfully executed with the use of tasam simulator.

Practical-5

Aim: Write a program to count the number of odds and evens from the given 8-bit array.

Software requirements: TASM Simulator

Logic:

assume cs:code,ds:data

data segment

n dw 1001h,8002h,0a003h,0b004h,5505h

d db ?

data ends

code segment

start : mov ax,data

mov ds,ax

mov cl,05h

lea si,n

again : mov ax,[si]

shl ax,1

jnc nocount

inc dl

nocount : inc si

shr ax,1

dec cl

jnz again

mov d,dl

mov ah,04ch

int 21h

code ends

end start

Output:

Conclusion:

The program of count the number of odds and evens from the given 8-bit array is successfully executed with the use of tasam simulator

Practical-6

Aim: Write a program to count the number of 1’s in 16-bit data.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

N DW 2306H

O DB ?

DATA ENDS

CODE SEGMENT

START : MOV AX,DATA

MOV DS,AX

MOV AX,N

MOV CL,0FH

MOV DL,00H

AGAIN  :  SHR AX,1

JNC NOCOUNT

INC DL

NOCOUNT :  LOOP AGAIN

MOV O,DL

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of count the number of 1’s in 16-bit data is successfully executed with the use of tasam simulator.

Practical-7

Aim: Write a program to find a factorial of a given number.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

N DW 3

ANS DW ?

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV AX,0001H

MOV CX,N

AGAIN:

MUL CX

DEC CX

JNZ AGAIN

MOV ANS,AX

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of find a factorial of a given number is successfully executed with the use of tasam simulator.

Practical-8

Aim: Write a program to find factors of the given number.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

N Dw 06H

ANS Dw ?

MSG DB '  FACTORS '

DATA ENDS

CODE SEGMENT

START: MOV AX,DATA

MOV DS,AX

MOV AH,9

LEA DX,MSG

MOV AX,06h

MOV CL,06h

lea di,ans

AGAIN:

mov ax,06h

DIV cl

CMP AH,00H

JZ FCT

DEC CL

JNZ AGAIN

JZ PEND

FCT:

mov [di],cl

inc di

INC BL

DEC CL

JNZ AGAIN

PEND:

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of find factors of the given number is successfully executed with the use of tasam simulator.

Practical-10

Aim: Write a program to find complete Fibonacci series till n th number. 

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

N DB 09H

A DB 01H

B DB 00H

MSG DB 'FIBONACCI','     ','SERIES'

SUM DB 01H

ANS DB ?

DATA ENDS

CODE SEGMENT

START:MOV AX,DATA

MOV DS,AX

MOV AH,9

LEA DX,MSG

MOV AL,A

MOV BL,B

MOV CL,N

MOV DL,SUM

lea si,ans

AGAIN:

CMP DL,CL

JNC AOUT

ADD AL,BL

CMP AL,CL

JNC AOUT

MOV DL,AL

mov [si],al

inc si

MOV AL,BL

MOV BL,DL

JMP AGAIN

AOUT:

MOV ANS,DL

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of find complete Fibonacci series till n th number is successfully executed with the use of tasam simulator.

Practical-11

Aim: Write a program to find a total number of negative elements in a given array.

Software requirements: TASM Simulator

Logic:

ASSUME CS:CODE,DS:DATA

DATA SEGMENT

N DW 1001H,8002H,0A003H,0B004H,5505H

D DB ?

DATA ENDS

CODE SEGMENT

START : MOV AX,DATA

MOV DS,AX

MOV CL,05H

LEA SI,N

AGAIN : MOV AX,[SI]

SHL AX,1

JNC NOCOUNT

INC DL

NOCOUNT : INC SI

SHR AX,1

DEC CL

JNZ AGAIN

MOV D,DL

MOV AH,04CH

INT 21H

CODE ENDS

END START

Output:

Conclusion:

The program of find a total number of negative elements in a given array is successfully executed with the use of tasam simulator.

system programming-quardple

void main()
{
char str[50],b,c,op[20],op1[20],op2[20];
int a=49,i,j,k=1,opi=0,op1i=0,op2i=0;
clrscr();
printf("enter string\n");
gets(str);
while(strlen(str)>=3)
{
i=0;
while(1)
{
if(str[i]>=33&&str[i]<=47)
{
op1[op1i++]=str[i-2];
str[i-2]=a++;
op[opi++]=str[i];
op2[op2i++]=str[i-1];
for(k=i;k<=strlen(str);k++)
str[k]=str[k+1];
str[k]=NULL;
for(k=i-1;k<=strlen(str);k++)
str[k]=str[k+1];
str[k]=NULL;
break;
}
else
{
i++;
}
}
}
op[opi]=NULL;
op1[op1i]=NULL;
op2[op2i]=NULL;
printf("\nTriple no. operator operator1 oprator2\n");
for(i=0;i<strlen(op);i++)
printf("%d\t\t%c\t%c\t%c\n",i+1,op[i],op1[i],op2[i]);
getch();
}

lexical analysis

//copyright this program is made by Jayesh Amin....
//This program is used to perform lexical Analysis....


//function for checking duplicate values

int check(char a[],char x,int z)
{
int y;
for(y=0;y<z;y++)
{
if(a[y]==x)
return 1;
}
return 0;
}

//function for printing lexical analysis

int print(char a[],char x1)
{
int q=0;
while(a[q]!=x1)
{
q++;
}
return q;
}

//main program starts

void main()
{

//variable declaration

int no,i,p,opi=0,keyi=0,idi=0,keyj=0,idp,opp,j,c,keyflag=1;
char stat[10][30],key[50],op[50],id[50],temp;
clrscr();

/* entering statements */

printf("\nHow many statements you want to enter???\n");
scanf("%d",&no);
for(i=0;i<no;i++)
{
printf("\nenter statement%d\n",i+1);
scanf("%s",stat[i]);
}
clrscr();
printf("\nyour statements are:\n\n");
for(i=0;i<no;i++)
{
printf("%s\n",stat[i]);
}
printf("\n");

//logic of the program starts

printf("your answer of lexical analysis is:\n\n");
for(i=0;i<no;i++)/*loop for no. of statements*/
{
for(j=0;j<strlen(stat[i]);j++)/*loop for statement length*/
{
/*if 1st char is identifier and after that ':' is there then do this*/
if((stat[i][j]>=97 && stat[i][j]<=122) && (stat[i][j+1]==':'))
{

if(check(id,temp,idi)==0)
{
id[idi]=stat[i][j];
idi++;
}
temp=stat[i][j];
idp=print(id,temp);
printf("(id#%d)",idp+1);
j++;
temp=stat[i][j];
if(check(op,temp,opi)==0)
{
op[opi]=stat[i][j];
opi++;
}
opp=print(op,temp);
printf("(op#%d)",opp+1);
j++;
while(stat[i][j]!=';')
{
key[keyi]=stat[i][j];
keyi++;
j++;
}
key[keyi]=' ';
keyi++;
printf("(key#%d)",keyflag);
keyflag++;
temp=stat[i][j];
if(check(op,temp,opi)==0)
{
op[opi]=stat[i][j];
opi++;
}
opp=print(op,temp);
printf("(op#%d)",opp+1);
break;
}

/*if 1st char is identifier and after that ',' is there then do this*/

else if((stat[i][j]>=97 && stat[i][j]<=122) && (stat[i][j+1]==','))
{
if(check(id,temp,idi)==0)
{
id[idi]=stat[i][j];
idi++;
}
temp=stat[i][j];
idp=print(id,temp);
printf("(id#%d)",idp+1);
j++;
temp=stat[i][j];
if(check(op,temp,opi)==0)
{
op[opi]=stat[i][j];
opi++;
}
opp=print(op,temp);
printf("(op#%d)",opp+1);
}

/*if 1st char is identifier and after that '=' is there then do this*/

else if((stat[i][j]>=97 && stat[i][j]<=122) && (stat[i][j+1]=='='))
{
temp=stat[i][j];
idp=print(id,temp);
printf("(id#%d)",idp+1);
j++;
temp=stat[i][j];
if(check(op,temp,opi)==0)
{
op[opi]=stat[i][j];
opi++;
}
opp=print(op,temp);
printf("(op#%d)",opp+1);
j++;
while(stat[i][j]!=';')
{
c=(int)stat[i][j];
if((c>=48 && c<=57)||(c>=65 && c<=90)||(c>=97 && c<=122))
{
temp=c;
idp=print(id,temp);
printf("(id#%d)",idp+1);
}
else
{
temp=stat[i][j];
if(check(op,temp,opi)==0)
{
op[opi]=stat[i][j];
opi++;
}
opp=print(op,temp);
printf("(op#%d)",opp+1);
}
j++;
}
temp=stat[i][j];
opp=print(op,temp);
printf("(op#%d)",opp+1);
break;
}
else
{
}
}
printf("\n");
}

op[opi]=NULL;
key[keyi]=NULL;
id[idi]=NULL;

/* To print identifiers,operators and keywords seperately */

printf("\nidentifiers are--->%s\n",id);
printf("\noperators are----->%s\n",op);
printf("\nkeywords are------>%s\n",key);
getch();
}

system programming-teminal and non terminal

void main()
{
/* variable declaration */

int no,i,p,m=0,n=0,j,flag;
char prod[10][20],ts[50],nts[50],c;
ts[0]=NULL,nts[0]=NULL;
clrscr();

/* entering production */

printf("\nhow many production you want to enter\n");
scanf("%d",&no);
for(i=0;i<no;i++)
{
printf("\nenter production%d\n",i+1);
scanf("%s",prod[i]);
}
for(i=0;i<no;i++)/*loop for no. of productions*/
{
for(j=0;j<strlen(prod[i]);j++)/*loop for production length*/
{
c=prod[i][j];
if(c>=97 && c<=122)/*check for terminal symbols*/
{
for(p=m;p>=0;p--)
{
if(ts[p]==c)
{
flag=1;
break;
}
else
flag=0;
}
if(flag==0)
{
ts[m]=c;
m++;
}
}
else if(c>=65 && c<=90)/*check for non terminal symbols*/
{
for(p=n;p>=0;p--)
{
if(nts[p]==c)
{
flag=1;
break;
}
else
flag=0;
}
if(flag==0)
{
nts[n]=c;
n++;
}
}
else
{
}
}
ts[m]=NULL;
nts[n]=NULL;
}

/*display final result*/

printf("\nterminl symbols are %s\n",ts);
printf("\nnon terminal symbols are %s",nts);
getch();
}

siedle method

#include
#include
#include

float ff (float x)
{
return (pow(x,2.0)-12.0);
}

void main()
{
int i=1;
float x1,x2,x3,e,f1,f2,f3;
clrscr();
printf("Enter value of tolerance : ");
scanf("%f",&e);
printf("enter x1 and x2 : ");
scanf("%f%f",&x1,&x2);
do{
printf("\n\n\n----------------------------------------------------");
printf("\n\n\nIteration %d",i);
printf("\n\n\nx1 = %f",x1);
printf("\t x2 = %f",x2);
f1=ff(x1);
f2=ff(x2);


x3=(x1*f2-x2*f1)/(f2-f1);
f3=ff(x3);
x1=x2;
x2=x3;
i++;
}while(fabs(f3)>e);
printf("\n\n\n------------------------------------------------\n\n");
printf("The root is : %f",x3);
getch();
}

jacobi mehod

#include
#include
#include
#include
void main()
{
float a[20][20],b,c,x[20],y[20];
int i,j,k,no,flag;
clrscr();
cout<<"how many eq. u want to enter?\n";
cin>>no;
for(i=1;i<=no;i++)
{
for(j=1;j<=no+1;j++)
{
printf("enter val. of a[%d][%d]",i,j);
cin>>a[i][j];
}
}
cout<<"matrix is\n";
for(i=1;i<=no;i++)
{
for(j=1;j<=no+1;j++)
{
cout<
cout<<"\t";
}
cout<<"\n";
}
for(k=1;k<=no;k++)
y[k]=0.0;
 for( k=1;k<=no;k++)
x[k]=0.0;
for(;;)
{
for(i=1;i<=no;i++)//i start
{
float temp=0.0;
for(j=1;j<=no;j++)
 {
if(i!=j)
{
temp+=a[i][j]*y[j];
x[i]=(a[i][no+1]-temp)/a[i][i];
y[i]=x[i];
}//if ends
 } //j ends
}   //i ends
int n=1;
for(int m=1;m<=no;m++)
{
if(x[m]-y[m]<=0.001)
n++;
}
if(n==no)
break;
}
for(j=1;j<=no;j++)
{
printf("\nx%d is\n",j);
cout<
}
cout<<"\n";
getch();
}

gauss elimination method

#include
#include
#include
void main()
{
int i=0,j=0,k=0,n=0;
float sum,a[100][100],temp,x[50];
clrscr();
printf("Enter the order of matrix : ");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
    for(j=1;j<=n+1;j++)
    {
    printf("\nEnter value of a[%d][%d] : ",i,j);
    scanf("%f",&a[i][j]);
    }
}
printf("\n\nEquations are : \n\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
{
printf("\t%f",a[i][j]);
}
printf("\n\n");
}
for(k=1;k<=n-1;k++)
{
for(j=k+1;j<=n;j++)
{
temp=a[j][k]/a[k][k];
for(i=k;i<=n+1;i++)
{
a[j][i]=a[j][i]-(temp)*a[k][i];
}
}
}
printf("\nNew Eq : \n\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
{
printf("\t%f",a[i][j]);
}
printf("\n");
}
x[n]=(a[n][n+1]/a[n][n]);
for(i=n-1;i>=0;i--)
{
sum=0;
for(j=i+1;j<=n;j++)
{
sum=sum+(a[i][j]*x[j]);
}
x[i]=(a[i][n+1]-sum)/a[i][i];
}
printf("\nRoots are : ");
for(i=1;i<=n;i++)
{
printf("\t%f",x[i]);
}
getch();
}