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  1. PO
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    <p>I am not familiar with the language syntax of <code>infix</code>, but you can certainly do a first pass validation check which simply verifies that all of the characters in the string are valid (i.e. acceptable characters = <code>A-Z</code>, <code>+</code>, <code>-</code>, <code>*</code>, <code>/</code>, <code>$</code>, <code>(</code> and <code>)</code>). Here is a Java program which checks for valid characters and also includes a function which checks for unbalanced (possibly nested) parentheses:</p> <pre class="lang-java prettyprint-override"><code>import java.util.regex.*; public class TEST { public static void main(String[] args) { String s = "A+B-(C/D)$(E+F)"; Pattern regex = Pattern.compile( "# Verify that a string contains only specified characters.\n" + "^ # Anchor to start of string\n" + "[A-Z+\\-*/$()]+ # Match one or more valid characters\n" + "$ # Anchor to end of string\n", Pattern.COMMENTS); Matcher m = regex.matcher(s); if (m.find()) { System.out.print("OK: String has only valid characters.\n"); } else { System.out.print("ERROR: String has invalid characters.\n"); } // Verify the string contains only balanced parentheses. if (checkParens(s)) { System.out.print("OK: String has no unbalanced parentheses.\n"); } else { System.out.print("ERROR: String has unbalanced parentheses.\n"); } } // Function checks is string contains any unbalanced parentheses. public static Boolean checkParens(String s) { Pattern regex = Pattern.compile("\\(([^()]*)\\)"); Matcher m = regex.matcher(s); // Loop removes matching nested parentheses from inside out. while (m.find()) { s = m.replaceFirst(m.group(1)); m.reset(s); } regex = Pattern.compile("[()]"); m = regex.matcher(s); // Check if there are any erroneous parentheses left over. if (m.find()) { return false; // String has unbalanced parens. } return true; // String has balanced parens. } } </code></pre> <p>This does not validate the grammar, but may be useful as a first test to filter out obviously bad strings.</p>
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    1. POValidating an infix notation possibly using regex
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    1. USridgerunner
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    1. POValidating an infix notation possibly using regex
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    1. VO
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    1. COfirst of all this is incredible inefficient. second, if you use it that way, why don't you replace [A-Z][+\\-*/$][A-Z] with A and \\([A-Z]\\) with A. So you end up with an empty string if the expression is valid
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      1. USduedl0r
 

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