evaluating postfix expression using stacks

[gau]

This program should first convert an infix expression to postfix expression then evaluate.

theres no problem in converting, the problem is in the evaluation part, here is the code.

#include <stdio.h>
#include <conio.h>
#include <string.h>
#include <ctype.h>

#define MAX 50

char stack[MAX] ;
char infix[MAX] ;
char postfix[MAX] ;
char eval[MAX] ;
char *s, *t ; /*pointers to input and output strings*/
int top; /*Stack top*/

/*Function Prototypes*/
void Initialize (void);
void SetExpression (char *);
char Pop (void );
void Push (char);
int priority (char);
void Convert (void);
int Evaluate(void);

void main( )
{
int m;

clrscr( ) ;
Initialize ( ) ;
printf ( "\nEnter an infix expression: " ) ;
gets ( infix ) ;
SetExpression (infix) ;
Convert( ) ;
printf ( "\nThe Postfix expression is: " ) ;
puts(postfix);
strcpy(eval,postfix);
m=Evaluate( );
printf("answer: %d", m );

getch( ) ;
}

void Initialize (void)
{
top = -1 ;/*Make stack empty*/
}

void SetExpression ( char *str )
{
s = str ;
t = postfix;
}

/* adds operator to the stack */
void Push ( char c )
{
if ( top == MAX - 1 )
printf ( "\nStack is full.\n" ) ;
else
{
top++ ;
stack[top] = c ;
}
}

/* pops an operator from the stack */
char Pop ( void )
{
if ( top == -1 ) /* Stack is empty*/
return -1 ;
else
{
char item = stack[top] ;
top-- ;
return item ;
}
}

int priority(char c)
{
if ( c == '*' || c == '/' || c == '%' )
return 2;
else if ( c == '+' || c == '-' )
return 1;
else
return 0;
}

/* converts the infix expr. to postfix form */
void Convert (void)
{
char x ;

while ( *( s ) )
{ /*Skip white spaces, if any*/
if ( *( s ) == ' ' || *( s ) == '\t' )
{
s++ ;
continue ;
}

if ( isdigit ( *( s ) ) )/*Operands*/
{
while ( isdigit ( *( s ) ) )
{
*( t ) = *( s ) ;
s++ ;
t++ ;
}
}
if ( *( s ) == '(' )/*Opening Parenthesis*/
{
Push ( *( s ) ) ;
s++ ;
}
if ( *( s ) == '*' || *( s ) == '+' || *( s ) == '/' || *( s ) == '%' || *( s ) == '-' ) /*operators*/
{
if ( top != -1 )
{
x = Pop ( ) ;
while ( priority ( x ) >= priority ( *( s ) ) )
{
*( t ) = x ;
t++ ;
x = Pop ( ) ;
}
Push( x ) ;
Push ( *( s ) ) ;
}
else Push( *( s ) ) ;
s++ ;
}
if ( *( s ) == ')' )/*Closing Parenthesis*/
{
x = Pop ( ) ;
while ( x != '(' )
{
*( t ) = x ;
t++ ;
x = Pop ( ) ;
}
s++ ;
}
}
while ( top != -1 )/*While stack is not empty*/
{
x = Pop ( ) ;
*( t ) = x ;
t++ ;
}
t++ ;
}

int Evaluate(void)
{
int i,l,a,b,q,z;
l=strlen(eval);
for(i=0;i<l;i++)
{
if ( isdigit ( eval ) )
{
Push(eval);
}
else if ( eval == '*' || eval == '+' || eval == '/' || eval == '%' || eval == '-' )
{
a = Pop ( );
b = Pop ( );

switch( eval )
{
case '+' : q=b+a; break;
case '-' : q=b-a; break;
case '*' : q=b*a; break;
case '/' : q=b/a; break;
}
Push ( q );
}
}
z = Pop ( );
return z;
}

I can't find whats wrong in the evaluation function, when i type 1+2+3 the answer is -106 T_T

can anyone help me?

 

[kpmsivachand]

In your program the problem occur in Evaluate fuction. In that

 a = Pop ( );
           b = Pop ( );

The pop function return the char according to prototype char Pop (void );
but the variable you are using in Evaluate function a and b are integer. So the ascii value can be taken.

Example:
Enter an infix expression:1+2
The Postfix expression is:21+
answer:99

Because: a=50,b=49 so answer:50+49=99

I hope it helps 8)

 

[kpmsivachand]

To avoid that add these lines in evaluate function....

a=pop();
b=pop();
a-=48;
b-=48;

 

[jepoy27]

i also want this program but i want it in c,

 

[jepoy27]

can anyone help me in project "evaluation postfix in c"........simple as possible thanks.......plsssssss....very urgent....

 

[jepoy27]

PLssssss.help me plsssssss....i running out of time....i really cant do it

 

[jepoy27]

can you help me with my project in school it is just the same of what you posted...evaluation of postfix but in C.......plsssss....i really need it very urgent thanksssss..........i expect you to help me....

 

[haritr]

please help me............Converting postfix expression into infix expression by using c progm