this is my code I want the opposite i want to convert a postfixexpression to infix expression

#include

#include

#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 ornot.

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 meanshigher 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

     {

    if(isOperator(infix[i])==1)

     {

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

  postfix += pop(st);

  push(st,infix[i]);

     }

     elseif(infix[i] == '(')

  push(st,infix[i]);

  

     elseif(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: \" <

  

  return 0;

}