0
package PJ2;
import java.util.*;

public class SimpleLispExpressionEvaluator
{
    // Current input Lisp expression
    private String inputExpr;

    // Main expression stack & current operation stack, see algorithm in evaluate()
    private Stack<Object> exprStack;
    private Stack<Double> currentOpStack;


    // default constructor
    // set inputExpr to "" 
    // create stack objects
    public SimpleLispExpressionEvaluator()
    {
    // add statements
        inputExpr = "";
        exprStack = new Stack<Object>();
        currentOpStack = new Stack<Double>();
    }

    // default constructor
    // set inputExpr to inputExpression 
    // create stack objects
    public SimpleLispExpressionEvaluator(String inputExpression) 
    {
    // add statements
        if(inputExpression == null)
        {
            throw new SimpleLispExpressionEvaluatorException();
        }

        inputExpr = inputExpression;
        exprStack = new Stack<Object>();
        currentOpStack = new Stack<Double>();

    }

    // set inputExpr to inputExpression 
    // clear stack objects
    public void reset(String inputExpression) 
    {
    // add statements
     if(inputExpression == null)
     {
         throw new SimpleLispExpressionEvaluatorException();
     }

     inputExpr = inputExpression;
     exprStack.clear();
     currentOpStack.clear();
    }

    // This function evaluate current operator with its operands
    // See complete algorithm in evaluate()
    //
    // Main Steps:
    //      Pop operands from exprStack and push them onto 
    //          currentOpStack until you find an operator
    //      Apply the operator to the operands on currentOpStack
    //          Push the result into exprStack
    //
    private void evaluateCurrentOperation()
    {
    // add statements
        String currOp;
        boolean numeric = true;
        do{
            currOp = (String.valueOf(exprStack.pop()));

            try{
                Double number = Double.parseDouble(currOp);
                currentOpStack.push(number);

            }catch(NumberFormatException nfe){
                numeric = false;
            }
        } while(numeric);


        double result;
        switch (currOp) {
            case "*":
                result = currentOpStack.pop();
                while(!currentOpStack.isEmpty()){
                    result *= currentOpStack.pop();
                }
                break;
            case "/":
                result = currentOpStack.pop();
                while(!currentOpStack.isEmpty()){
                    result /= currentOpStack.pop();
                }
                break;
            case "+":
                result = currentOpStack.pop();
                while(!currentOpStack.isEmpty()){
                    result += currentOpStack.pop();
                }
                break;
            case "-":
                result = currentOpStack.pop();
                while(!currentOpStack.isEmpty()){
                    result -= currentOpStack.pop();
                }
                break;

            default:
                result = currentOpStack.pop();
                break;
        }

        exprStack.push(result);

    }

    /**
     * This function evaluates current Lisp expression in inputExpr
     * It return result of the expression 
     *
     * The algorithm:  
     *
     * Step 1   Scan the tokens in the string.
     * Step 2       If you see an operand, push operand object onto the exprStack
     * Step 3           If you see "(", next token should be an operator
     * Step 4       If you see an operator, push operator object onto the exprStack
     * Step 5       If you see ")", do steps 6,7,8 in evaluateCurrentOperation() :
     * Step 6           Pop operands and push them onto currentOpStack 
     *                  until you find an operator
     * Step 7           Apply the operator to the operands on currentOpStack
     * Step 8           Push the result into exprStack
     * Step 9    If you run out of tokens, the value on the top of exprStack is
     *           is the result of the expression.
     */
    public double evaluate()
    {
        // only outline is given...
        // you need to add statements/local variables
        // you may delete or modify any statements in this method

        // use scanner to tokenize inputExpr
        Scanner inputExprScanner = new Scanner(inputExpr);

        // Use zero or more white space as delimiter,
        // which breaks the string into single character tokens
        inputExprScanner = inputExprScanner.useDelimiter("\\s*");

        // Step 1: Scan the tokens in the string.
        while (inputExprScanner.hasNext())
        {

            // Step 2: If you see an operand, push operand object onto the exprStack
            if (inputExprScanner.hasNextInt())
            {
                // This force scanner to grab all of the digits
                // Otherwise, it will just get one char
                String dataString = inputExprScanner.findInLine("\\d+");

        // more ...
                exprStack.push(Double.parseDouble(dataString));
            }
            else
            {
                // Get next token, only one char in string token
                String aToken = inputExprScanner.next();
                char item = aToken.charAt(0);

                switch (item)
                {
                case '(':
                    break;

                case ')':
                    evaluateCurrentOperation();
                    break; 

                case'*':
                    exprStack.push("*");
                    break;

                case'+':
                    exprStack.push("+");
                    break; 

                case'/':
                    exprStack.push("/");
                    break;

                case'-':
                    exprStack.push("-");
                    break;  

                // Step 3: If you see "(", next token shoube an operator
                // Step 4: If you see an operator, push operator object onto the exprStack
                // Step 5: If you see ")"  do steps 6,7,8 in evaluateCurrentOperation() :
                    default:  // error
                        throw new SimpleLispExpressionEvaluatorException(item + " is not a legal expression operator");
                } // end switch
            } // end else
        } // end while

        // Step 9: If you run out of tokens, the value on the top of exprStack is
        //         is the result of the expression.
        //
        //         return result
        return Double.parseDouble(String.valueOf(exprStack.pop()));
    }


    //=====================================================================

    // This static method is used by main() only
    private static void evaluateExprTest(String s, SimpleLispExpressionEvaluator expr)
    {
        Double result;
        System.out.println("Expression " + s);
    expr.reset(s);
        result = expr.evaluate();
        System.out.printf("Result %.2f\n", result);
        System.out.println("-----------------------------");
    }

    // define few test cases, exception may happen 
    public static void main (String args[])
    {
        SimpleLispExpressionEvaluator expr= new SimpleLispExpressionEvaluator();
        String test1 = "(+ (- 6) (* 2 3 4) (/ (+ 3) (* 1) (- 2 3 1)))";
        String test2 = "(+ (- 632) (* 21 3 4) (/ (+ 32) (* 1) (- 21 3 1)))";
        String test3 = "(+ (/ 2) (* 2) (/ (+ 1) (+ 1) (- 2 1 )))";
        String test4 = "(+ (/2))";
        String test5 = "(+ (/2 3 0))";
        String test6 = "(+ (/ 2) (* 2) (/ (+ 1) (+ 3) (- 2 1 ))))";
        String test7 = "( + 100 )";
        String test8 = "( + 100 5 5 5 )";
        String test9 = "(*(/100 5)4      10)";
        String test10 ="( + 100 )";
        String test11 ="( + ( - 6 ) ( * 2 3 4 ) ( / ( + 3 ) ( * 1 ) ( - 2 3 1 ) ) )";
        String test12 ="( + ( - 632 ) ( * 21 3 4 ) ( / ( + 32 ) ( * 1 ) ( - 21 3 1 ) ) )";
        String test13 ="";
        String test14 ="";

    evaluateExprTest(test1, expr);
    evaluateExprTest(test2, expr);
    evaluateExprTest(test3, expr);
    evaluateExprTest(test4, expr);
    evaluateExprTest(test5, expr);
    evaluateExprTest(test6, expr);
    evaluateExprTest(test7, expr);
    evaluateExprTest(test8, expr);
    evaluateExprTest(test9, expr);
    evaluateExprTest(test10, expr);
    evaluateExprTest(test11, expr);
    evaluateExprTest(test12, expr);
    //evaluateExprTest(test13, expr);
    //evaluateExprTest(test14, expr);
    }
}

私が実行すると、3,4,5,7,8,9,10 のみが正しく動作することがテストされます: http://joeganley.com/code/jslisp.html

Expression (+ (- 6) (* 2 3 4) (/ (+ 3) (* 1) (- 2 3 1)))
Result 28.50
-----------------------------
Expression (+ (- 632) (* 21 3 4) (/ (+ 32) (* 1) (- 21 3 1)))
Result 885.88
-----------------------------
Expression (+ (/ 2) (* 2) (/ (+ 1) (+ 1) (- 2 1 )))
Result 5.00
-----------------------------
Expression (+ (/2))
Result 2.00
-----------------------------
Expression (+ (/2 3 0))
Result Infinity
-----------------------------
Expression ( + 100 )
Result 100.00
-----------------------------
Expression ( + 100 5 5 5 )
Result 115.00
-----------------------------
Expression (*(/100 5)4      10)
Result 800.00
-----------------------------
Expression ( + 100 )
Result 100.00
-----------------------------
Expression ( + ( - 6 ) ( * 2 3 4 ) ( / ( + 3 ) ( * 1 ) ( - 2 3 1 ) ) )
Result 28.50
-----------------------------
Expression ( + ( - 632 ) ( * 21 3 4 ) ( / ( + 32 ) ( * 1 ) ( - 21 3 1 ) ) )
Result 885.88
-----------------------------

これらの結果は、例外が発生して実行が停止するため、テスト 6 を削除した場合にのみ発生します。私がどこで間違ったのか、何か考えはありますか?

6 がコメントアウトされていない場合のエラー:

Exception in thread "main" java.util.EmptyStackException
    at java.util.Stack.peek(Unknown Source)
    at java.util.Stack.pop(Unknown Source)
    at PJ2.SimpleLispExpressionEvaluator.evaluateCurrentOperation(SimpleLispExpressionEvaluator.java:143)
    at PJ2.SimpleLispExpressionEvaluator.evaluate(SimpleLispExpressionEvaluator.java:248)
    at PJ2.SimpleLispExpressionEvaluator.evaluateExprTest(SimpleLispExpressionEvaluator.java:292)
    at PJ2.SimpleLispExpressionEvaluator.main(SimpleLispExpressionEvaluator.java:321)
4

1 に答える 1

5

それは本当の Lisp ではありません。それは接頭辞形式の算術式です。Lispの約0.1%。Lisp はもう少し: 関数、シンボル、コンス セル、リスト操作、...

テスト式 6 には、最後に余分な括弧があります。

一般に、構文エラーを簡単に検出できるようにコードを開発する必要があります。つまり、エラー検出、エラー処理、およびエラー報告です。典型的な Lisp システム (たとえば、SBCL、CLISP など) は、余分な括弧を平易な英語でユーザーに報告します。

于 2013-04-09T07:00:02.857 に答える