Question

i want similar for this code to solve two questions :

1- Write a program to convert a postfix expression to infix expression

2-Write a program to convert an infix expression to prefix expression

each question in separate code ( so will be two codes )

#include <iostream>

#include <string>

#define SIZE 50

using namespace std;

// structure to represent a stack

struct Stack {

  char s[SIZE];

  int top;

};

void push(Stack *st, char c) {

  st->top++;

  st->s[st->top] = c;

}

char pop(Stack *st) {

  char c = st->s[st->top];

  st->top--;

  //(A+B)*(C+D)

  return c;

}

/* function to check whether a character is an operator or not.

this function returns 1 if character is operator else returns 0 */

int isOperator(char c) {

  switch (c)

  {

  case '^':

  case '+':

  case '-':

  case '*':

  case '/':

  case '%':

  return 1;

  default:

  return 0;

  }

}

/* function to assign precedence to operator.

In this function we assume that higher integer value means higher precedence */

int precd(char c) {

  switch (c)

  {

  case '^':

  return 3;

  case '*':

  case '/':

  case '%':

  return 2;

  case '+':

  case '-':

  return 1;

  default:

  return 0;

  }

}

//function to convert infix expression to postfix expression

string infixToPostfix(Stack *st, string infix) {

  string postfix = "";

  for(int i=0; i<infix.length(); i++)

     {

     if(isOperator(infix[i])==1)

     {

  while(st->top!=-1 && precd(st->s[st->top]) >= precd(infix[i]))

  postfix += pop(st);

  push(st,infix[i]);

     }

     else if(infix[i] == '(')

  push(st,infix[i]);

  

     else if(infix[i] == ')')

     {

  while(st->top!=-1 && st->s[st->top] != '(')

  postfix += pop(st);

  pop(st);

     }

     else

  postfix += infix[i];

  

  }

  while(st->top != -1)

  postfix += pop(st);

return postfix;

}

int main() {

  Stack st;

  st.top = -1;

  string infix;

  cout << "Enter an infix expression: ";

  getline(cin, infix);

  cout << "Postfix expression is: " << infixToPostfix(&st, infix);

  

  return 0;

}

Answer 1

1)//function to convert postfix expression to infix expression

string PostToInfix(Stack *st, string postfix) {

  string infix = "";

  for(int i=0; i< postfix.length(); i++)

     {

//If postfix[i]) is operator follow the given lines

//precedence is not needed in postfix expressions

     if(isOperator( postfix[i])==1)

     {

  while(st->top!=-1)

string op1;

string op2;

op1=st->top;

pop(st);

op2=st->top;

pop(st);

infix='('+op2+postfix[i]+op1+')';

  push(st,infix[i]);

     }

else

//this condition runs if postfix[i] is an operand

   push(st,postfix[i]);

    }

return infix;

}

int main() {

  Stack st;

  st.top = -1;

  string postfix;

cout << "Enter an postfix expression: ";

  getline(cin, postfix);

  cout << "Infix expression is: " << PostToInfix(&st,postfix);

  return 0;

}

2)//function to convert infix expression to prefix expression

string infixToprefix(Stack *st, string infix) {

  string prefix = "";

reverse(infix.begin(),infix.end());

for(int i=0; i<infix [i].length(); i++)

{

if(infix[i] == '(')

infix[i] == ')';

if(infix[i] == ')')

infix[i] == '(';

}

  for(int i=0; i<infix [i].length(); i++)

     {

     if(isOperator(infix[i])==1)

     {

if(st->top!=-1 && precd(infix[i])>=precd(st->s[st->top]))

push(st,infix[i]);

else if (precd(infix[i])==precd(st->s[st->top])&& (infix[i]=='^'))

{ prefix += pop(st);

pop(st);

}

else

   push(st,infix[i]);

     }

     

for(int i=0; i<infix [i].length(); i++)

{

if(infix[i] == ')')

{

while(st->top!=-1 && st->s[st->top] != '(')

{   prefix += infix[i];

pop(st);

}

}

else if(infix[i] == '(')

  push(st,infix[i])

return prefix;

}

}

int main() {

  Stack st;

  st.top = -1;

  string infix;

  cout << "Enter an infix expression: ";

  getline(cin, infix);

  cout << "Prefix expression is: " << infixToprefix(&st, infix);

  

  return 0;

}