Home Backend Development C++ How can the Shunting-yard algorithm be used to convert a mathematical expression string into a tree structure in C ?

How can the Shunting-yard algorithm be used to convert a mathematical expression string into a tree structure in C ?

Oct 30, 2024 am 08:37 AM

How can the Shunting-yard algorithm be used to convert a mathematical expression string into a tree structure in C  ?

Parsing Mathematical Expressions in C Using the Shunting-yard Algorithm

In the realm of programming, expressing complex mathematical calculations in code often requires parsing text strings into an internal tree representation. This facilitates subsequent evaluation and manipulation of these expressions.

Consider the task of parsing the following math expression string: "(a b)c-(d-e)f/g". The goal is to construct a tree structure using C classes to represent this expression.

Using the Shunting-yard Algorithm

The Shunting-yard algorithm provides an effective strategy for parsing mathematical expressions into trees. This algorithm operates in two phases:

  1. Tokenize the Expression: Break the string into its individual tokens, including operands (e.g., "a", "b"), operators (e.g., " ", "-"), and parentheses.
  2. Build the Tree: Use two stacks: an operator stack and an output stack. Process the tokens one at a time:

    • If the token is an operand, push it onto the output stack.
    • If the token is an operator, pop operators from the operator stack until you encounter a lower-precedence operator or an open parenthesis. Push the current operator onto the operator stack.
    • If the token is an open parenthesis, push it onto the operator stack.
    • If the token is a closed parenthesis, pop operators from the operator stack until you encounter the corresponding open parenthesis. Push the resulting subexpression onto the output stack.

Defining the Tree Structure

To represent the tree structure, define the following C classes:

  • Exp (base class)
  • Term (inheriting from Exp) for operands
  • Node (inheriting from Exp) for operators

Example Parsing Process

For the expression "(a b)c-(d-e)f/g", the parsing process would proceed as follows:

Operator Stack | Output Stack
--------------|--------------
               | a b
+             | a b +
               | a b + c
*             | a b + c *
               | a b + c * d
-             | a b + c * d -
               | a b + c * d - e
               | a b + c * (d - e)
*             | a b + c * (d - e) f
               | a b + c * (d - e) f /
               | (a + b) * c - (d - e) * f / g
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The resulting tree structure would have the following form:

            *
            / \
       (a + b) * (d - e)
            / \
           /   \
          c     / \
               f / g
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