converting infix to postfix (assignment operator included) using c++ - c++

Question

I was trying to convert to an infix expression to postfix in c++. I have done coding with simple operators /,*,+,- but I am confused about the logic for assignment operator(=). I know it must be given the lowest precedence but for input like A=B=4 or A=(B=2)*2, how can I decide the precedence. My code is written below. (without implementation for assignment operator).
#include <iostream>
#include <stack>
#include <string>
using namespace std;
// Simply determine if character is one of the four standard operators.
bool isOperator(char character) {
if (character == '+' || character == '-' || character == '*' || character == '/' || character=='=') {
return true;
}
return false;
}
// If the character is not an operator or a parenthesis, then it is assumed to be an operand.
bool isOperand(char character) {
if (!isOperator(character) && character != '(' && character != ')') {
return true;
}
return false;
}
// Compare operator precedence of main operators.
// Return 0 if equal, -1 if op2 is less than op1, and 1 if op2 is greater than op1.
int compareOperators(char op1, char op2) {
if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) { return -1; }
else if ((op1 == '+' || op1 == '-') && (op2 == '*' || op2 == '/')) { return 1; }
return 0;
}
int main()
{
// Empty character stack and blank postfix string.
stack<char> opStack;
string postFixString = "";
char input[100];
// Collect input
cout << "Enter an expression: ";
cin >> input;
// Get a pointer to our character array.
char *cPtr = input;
// Loop through the array (one character at a time) until we reach the end of the string.
while (*cPtr != '\0') {
// If operand, simply add it to our postfix string.
// If it is an operator, pop operators off our stack until it is empty, an open parenthesis or an operator with less than or equal precedence.
if (isOperand(*cPtr)) { postFixString += *cPtr; }
else if (isOperator(*cPtr)) {
while (!opStack.empty() && opStack.top() != '(' && compareOperators(opStack.top(), *cPtr) <= 0) {
postFixString += opStack.top();
opStack.pop();
}
opStack.push(*cPtr);
}
// Simply push all open parenthesis onto our stack
// When we reach a closing one, start popping off operators until we run into the opening parenthesis.
else if (*cPtr == '(') { opStack.push(*cPtr); }
else if (*cPtr == ')') {
while (!opStack.empty()) {
if (opStack.top() == '(') { opStack.pop(); break; }
postFixString += opStack.top();
opStack.pop();
}
}
// Advance our pointer to next character in string.
cPtr++;
}
// After the input expression has been ran through, if there is any remaining operators left on the stack
// pop them off and put them onto the postfix string.
while (!opStack.empty()) {
postFixString += opStack.top();
opStack.pop();
}
// Show the postfix string at the end.
cout << "Postfix is: " << postFixString << endl;
return 0;
}

Answer 1

Try giving it the same precedence as the operands, and see what you get:
// Simply determine if character is one of the four standard operators.
bool isOperator(char character) {
if (character == '+' || character == '-' || character == '*' || character == '/') {
//|| character=='='
return true;
}
return false;
}
// If the character is not an operator or a parenthesis, then it is assumed to be an operand.
bool isOperand(char character) {
//for =, treat it wth the same rules as the operands; push to output
if (character == '='){
return true;
}
if (!isOperator(character) && character != '(' && character != ')') {
return true;
}
return false;
}