Question

(In C)

Note: Can you create an example code of these tasks. use any variables you wish to use.

postfix expressions: (each individual text (T), (F), (NOT), etc is a token)

F
T NOT
T
F NOT
T
T T AND
F
T F NAND

(1) Create stack s.

(2) For each token, x, in the postfix expression:

If x is T or F push it into the stack s. (T = true, F = false)

Else if x is a unary operator

i pop an operand, op1, from s

ii compute x op1 (see unary table below)

   iii push the result into s

Else if x is a binary operator

i pop an operand, op2, from s

ii pop an operand, op1, from s

iii compute op1 op2 x iv push the result into s

op1 NOT

!op1

Binary operations

op1 op2 AND	op1 && op2
op1 op2 NAND	!(op1 && op2)
op1 op2 OR	op1 || op2
op1 op2 NOR	!(op1 || op2)
op1 op2 XOR	op1 ! = op2
op1 op2 COND	!op1 || op2
op1 op2 BICOND	op1 == op2

Answer 1

# your code goes here

static int evaluatePostfix(String exp)
{ //We are assuming True as 1 and False as 0 hence creating a stack of Integer type
//create a stack
Stack<Integer> stack=new Stack<Integer>();
//expression is taken as string ans exp.length() is giving the length of string
// Scan all characters one by one
for(int i=0;i<exp.length();i++)
{
char c=exp.charAt(i); //giving character present at position i
  
// If the scanned character is an operand (number here),
// push it to the stack.
if(Character.isDigit(c))
stack.push(c - '0');
  
// If the scanned character is an unary operator, pop one
// element from stack apply the operator
  
else if(c == "!"){
   int val = stack.pop();
   val = !val;
   stack.push(val);
}
  
/ If the scanned character is a binary operator, pop two
// elements from stack apply the operator
else
{
int val1 = stack.pop();
int val2 = stack.pop();
  
switch(c)
{
case '&&':
stack.push(val2&&val1);
break;
  
case '||':
stack.push(val2 || val1);
break;
  
case '^':
stack.push(val2^val1);
break;
  
case '==':
stack.push(val2==val1);
break;
}
}
}
return stack.pop();   
}