Write a program in C language to convert infix expression to postfix expression?

Answer 1

#

/**************************//**********cReDo**********//*****mchinmay@live.com***///C PROGRAM TO CONVERT GIVEN VALID INFIX EXPRESSION INTO POSTFIX EXPRESSION USING STACKS.#include#include#include#define MAX 20char stack[MAX];int top=-1;char pop();void push(char item);int prcd(char symbol){switch(symbol){case '+':case '-':return 2;break;case '*':case '/':return 4;break;case '^':case '$':return 6;break;case '(':case ')':case '#':return 1;break;}}int isoperator(char symbol){switch(symbol){case '+':case '-':case '*':case '/':case '^':case '$':case '(':case ')':return 1;break;default:return 0;}}void convertip(char infix[],char postfix[]){int i,symbol,j=0;stack[++top]='#';for(i=0;i{symbol=infix[i];if(isoperator(symbol)==0){postfix[j]=symbol;j++;}else{if(symbol=='(')push(symbol);else if(symbol==')'){while(stack[top]!='('){postfix[j]=pop();j++;}pop();//pop out (.}else{if(prcd(symbol)>prcd(stack[top]))push(symbol);else{while(prcd(symbol)<=prcd(stack[top])){postfix[j]=pop();j++;}push(symbol);}//end of else.}//end of else.}//end of else.}//end of for.while(stack[top]!='#'){postfix[j]=pop();j++;}postfix[j]='\0';//null terminate string.}void main(){char infix[20],postfix[20];clrscr();printf("Enter the valid infix string:\n");gets(infix);convertip(infix,postfix);printf("The corresponding postfix string is:\n");puts(postfix);getch();}void push(char item){top++;stack[top]=item;}char pop(){char a;a=stack[top];top--;return a;}//mail me: mchinmay@live.com

Answer 2

/*A VERY SIMPLE PROGRAM TO EVALUATE POSTFIX EXPRESSION USING STACK */

#include<iostream.h>

#include<conio.h>

#include<string.h>

void main()

{

int *stack;

char *p;

int top=0,i=0;

clrscr();

cout<<"WELCOME TO THE PROGRAM MADE BY SAURABH GNDU JALANDHAR "<<endl;

cout<<""<<endl;

cout<<"ENTER THE POSTFIX EXPRESSION "<<endl;

cin>>p;

int c = strlen(p);

while(i<c)

{

if(!( (p[i]>=42) && (p[i]<=47)) )

{

top = top +1;

stack[top] = p[i];

stack[top] = stack[top] - 48;

//cout<<stack[top]<<endl;

}

else

{

int temp1 , temp2;

temp1 = stack[top];

//cout<<temp1<<endl;

top--;

temp2 = stack[top];

//cout<<temp2<<endl;

top--;

int result;

switch(p[i])

{

case '+':

result = temp2 + temp1;

//cout<<result<<endl;

break;

case '-':

result = temp2 - temp1;

break;

case '*':

result = temp2 * temp1;

break;

case '/':

result = temp2 /temp1;

break;

}

top++;

stack[top]= result;

}

i++;

}

top=1;

cout<<stack[top]<<endl;

getch();

}

Answer 3

#include
#include
#include
struct stack
{
int top;
char item[100];
};
int empty(struct stack *s)
{
if (s->top==-1)
return 1;
else
return 0;
}
char popandtest(struct stack *ps)
{
return (ps->item[ps->top--]);
}
void pushandtest(struct stack *ps,char symb)
{
ps->top=ps->top+1;;
ps->item[ps->top]=symb;
}
int ISDIGIT(char symb)
{
if (symb>='0' && symb<='9')
{
return 1;
}
else
{
return 0;
}
}
double power(double op1,double op2)
{
int i;
double op11;
op11=op1;
for (i=1;i{
op1=op1*op11;
}
return op1;
}
double oper(char symb,double op1,double op2)
{

switch (symb)
{
case '+':

return (op1+op2);
break;
case '-':

return (op1-op2);
break;
case '*':

return (op1*op2);
break;
case '/':

return (op1/op2);
break;
case '$':

return (power(op1,op2));
break;
default:
printf("invalid operator\n");
return 0;
break;
}
}
double evalpre()
{
struct stack ps;
int count=0;

ps.top=-1;
int len;
int i,Break;
char string[100];
char symbol,top1,top2;
double op1,op2,value;
printf("Enter expression in prefix form values\n");
scanf("%s",string);
len=strlen(string);
for (i=0;i{
symbol=string[i];

if(ps.top>=1)
if(ISDIGIT(top1=ps.item[ps.top])==1 && ISDIGIT(top2=ps.item[ps.top-1])==1)
{
Break=1;
char symbol2;
while(Break!=0)
{
op2=(double)popandtest(&ps)-'0';
op1=(double)popandtest(&ps)-'0';
symbol2=popandtest(&ps);
value=oper(symbol2,op1,op2);
pushandtest(&ps,(char)value);
top1=ps.item[ps.top];
top2==ps.item[ps.top-1];
if(ISDIGIT(top1)==1 && ISDIGIT(top2)==1)
Break=1;
else
Break=0;
}
pushandtest(&ps,symbol);
}



else
pushandtest(&ps,symbol);

}
char symbol2;
while(ps.top!=0)
{
op2=(double)popandtest(&ps)-'0';
op1=(double)popandtest(&ps)-'0';
symbol2=popandtest(&ps);
value=oper(symbol2,op1,op2);
pushandtest(&ps,(char)value);
}
printf("top=%d\n",ps.top);
value=((double)(popandtest(&ps)-'0'));

return (value);
}
int main()
{
double val;
val=evalpre();
printf("value = %f\n",val);
return 0;
}





what problem does have this program?

Answer 4

/*Postfix->Evaluated value */

#include

#include

#include

#define MAXCOLS 80

#define TRUE 1

#define FALSE 0

struct stack{

int top;

double items[MAXCOLS];

};

double eval(char[]);

double pop(struct stack *);

void push(struct stack *,double);

int empty(struct stack *);

int isdigit(char);

double oper(int,double,double);

void main()

{

char expr[MAXCOLS];

int position =0;

while((expr[position++] = getchar()) != '\n') ;

expr[--position] = '\0';

printf("%s%s", " the original postfix expression is ",expr);

getchar();

printf("\n %f ", eval(expr));

getchar();

}

double pop(struct stack *ps)

{

double x;

if (ps->top==-1)

{

printf("%s","stack underflow");

exit(1);

}

else

{

x=ps->items[ps->top];

ps->top=ps->top - 1;

}

return x;

}

void push(struct stack *ps, double x)

{

if (ps->top == MAXCOLS-1)

{

printf("%s","stack overflow ");

exit(1);

}

else

{

ps->top=ps->top + 1;

ps->items[ps->top]=x;

}

}

double eval(char expr[])

{

int c , position;

double opnd1, opnd2, value ;

struct stack s;

s.top=-1;

for(position =0; (c=expr[position])!='\0'; position++)

if (isdigit(c))

push(&s,(double)(c-'0'));

else

{

opnd2=pop(&s);

opnd1=pop(&s);

value=oper(c ,opnd1 ,opnd2);

push(&s,value);

}

return(pop(&s));

}

int isdigit(char symb)

{

return ( symb >= '0' && symb <= '9');

}

double oper(int symb , double op1 , double op2)

{

switch (symb)

{

case '+' :return (op1 + op2);

case '- ' :return (op1 - op2);

case '*' :return (op1 * op2);

case '/' :return (op1 / op2);

case '$' :return (pow(op1 , op2));

default : printf("%s", "illegal operation ");

exit(1);

}

}

/*

This program is prepared for eveluating a given postfixexpression and when you run this program postfix expression must be given as input string and program will generate a numeric real number that is called as evaluated value.

If you give input string as the output of previous program 9 1 4 2 / + - 2 3 * / Program will generate evaluated value as 1.0

*/

Answer 5

/**************************//**********cReDo**********//*****mchinmay@live.com***///C PROGRAM TO CONVERT GIVEN VALID INFIX EXPRESSION INTO POSTFIX EXPRESSION USING STACKS.#include#include#include#define MAX 20char stack[MAX];int top=-1;char pop();void push(char item);int prcd(char symbol){switch(symbol){case '+':case '-':return 2;break;case '*':case '/':return 4;break;case '^':case '$':return 6;break;case '(':case ')':case '#':return 1;break;}}int isoperator(char symbol){switch(symbol){case '+':case '-':case '*':case '/':case '^':case '$':case '(':case ')':return 1;break;default:return 0;}}void convertip(char infix[],char postfix[]){int i,symbol,j=0;stack[++top]='#';for(i=0;i{symbol=infix[i];if(isoperator(symbol)==0){postfix[j]=symbol;j++;}else{if(symbol=='(')push(symbol);else if(symbol==')'){while(stack[top]!='('){postfix[j]=pop();j++;}pop();//pop out (.}else{if(prcd(symbol)>prcd(stack[top]))push(symbol);else{while(prcd(symbol)<=prcd(stack[top])){postfix[j]=pop();j++;}push(symbol);}//end of else.}//end of else.}//end of else.}//end of for.while(stack[top]!='#'){postfix[j]=pop();j++;}postfix[j]='\0';//null terminate string.}void main(){char infix[20],postfix[20];clrscr();printf("Enter the valid infix string:\n");gets(infix);convertip(infix,postfix);printf("The corresponding postfix string is:\n");puts(postfix);getch();}void push(char item){top++;stack[top]=item;}char pop(){char a;a=stack[top];top--;return a;}//mail me: mchinmay@live.com

Answer 6

Yes, you should.

Answer 7

No