title: How to evaluate an infix expression in just one scan using stacks?
body: <p>I want to know if there is a way to solve infix expressions in a single pass using 2 stacks?
  The stacks can be one for operator and the other for operands...</p>
  
  <p>The standard way to solve by shunt-yard algorithm is to convert the infix expression to postfix(reverse polish) and then solve. I don't want to convert the expression first to postfix.</p>
  
  <p>If the expression is like <code>2*3-(6+5)+8</code>, how to solve?</p>