Solving the Infix to Postfix Conversion Problem: Handling Parentheses Issues in C

Explore a concise solution to the infix to postfix conversion problem in C. Learn how to eliminate lingering parentheses from fully enclosed expressions for successful algorithm execution!
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Understanding the Infix to Postfix Conversion Problem

In the realm of data structures, one common task involves converting infix expressions (like (2 + 2)) into postfix notation (which would be 2 2 + ). While the algorithm to perform this conversion is relatively simple, there can be specific pitfalls, particularly when dealing with parentheses. Often, students encounter unique cases where an extra left parenthesis ends up in the output, such as in the case of fully enclosed expressions.

Imagine this situation: when you convert a simple expression like (2 + 2), you expect the result to be 2 2 + . However, due to a lingering left parenthesis, the output becomes 2 2 + (—an incorrect representation. The question then arises: why does this happen and how can we rectify it?

The Problem: Extraneous Parentheses

The main issue is that during the transfer of nodes from your stack to the output queue, the next pointer of the transferred node still points to the original node’s successor in the stack. Thus, when the stack is subsequently modified, it can inadvertently affect the queue, leading to unexpected results.

This situation becomes particularly problematic for cases with fully enclosed expressions, such as (2 + 2), where it seems as though the left parenthesis is being mistakenly included in the output.

Key Findings

Node Transfer Issue: The lingering left parenthesis will remain in the result if its pointer is still active during the enqueueing process.

Scope of Error: The error occurs predominantly during the pop/enqueue sequence of the conversion logic.

The Solution: Clearing the Pointer

To address the issue effectively, we need to ensure that the next pointer of each node being transferred to the queue is properly cleared. Here's the step-by-step breakdown of the solution:

Step 1: Modify the Enqueue Function

In the enqueue function of your implementation, you’ll need to add a simple assignment to clear the next field of the node being enqueued:

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By adding np->next = NULL;, you eliminate any ambiguity concerning the next node that the queue might erroneously reference.

Step 2: Checking the Push Function

While the push function itself may not contribute directly to the lingering parenthesis issue, it’s essential to ensure that it does not propagate problems. Typically, it follows a pattern of setting the next pointer to the current stack top. As a convention, you might want to check its implementation to ensure that this part also is running smoothly.

Conclusion

With this modification, you should now find that the extra left parentheses no longer appear in cases of fully enclosed expressions. Thus, when you run your program, expressions like (2 + 2) should successfully convert to 2 2 + without lingering characters. This small change can have a considerable impact on ensuring your algorithm runs correctly.

Now, as you continue to refine your data structures skills, remember that even a simple pointer assignment can be a game-changer in achieving expected results. It’s all about attention to detail in coding!