JDK14/Java14源码在线阅读

/*
 * Copyright (c) 1999, 2008, Oracle and/or its affiliates. All rights reserved.
 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
 *
 * This code is free software; you can redistribute it and/or modify it
 * under the terms of the GNU General Public License version 2 only, as
 * published by the Free Software Foundation.  Oracle designates this
 * particular file as subject to the "Classpath" exception as provided
 * by Oracle in the LICENSE file that accompanied this code.
 *
 * This code is distributed in the hope that it will be useful, but WITHOUT
 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
 * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License
 * version 2 for more details (a copy is included in the LICENSE file that
 * accompanied this code).
 *
 * You should have received a copy of the GNU General Public License version
 * 2 along with this work; if not, write to the Free Software Foundation,
 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
 *
 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
 * or visit www.oracle.com if you need additional information or have any
 * questions.
 */

package javax.management;


/**
 * This class is used by the query-building mechanism to represent binary
 * operations.
 * @serial include
 *
 * @since 1.5
 */
class BinaryOpValueExp extends QueryEval implements ValueExp {

    /* Serial version */
    private static final long serialVersionUID = 1216286847881456786L;

    /**
     * @serial The operator
     */
    private int op;

    /**
     * @serial The first value
     */
    private ValueExp exp1;

    /**
     * @serial The second value
     */
    private ValueExp exp2;


    /**
     * Basic Constructor.
     */
    public BinaryOpValueExp() {
    }

    /**
     * Creates a new BinaryOpValueExp using operator o applied on v1 and
     * v2 values.
     */
    public BinaryOpValueExp(int o, ValueExp v1, ValueExp v2) {
        op   = o;
        exp1 = v1;
        exp2 = v2;
    }


    /**
     * Returns the operator of the value expression.
     */
    public int getOperator()  {
        return op;
    }

    /**
     * Returns the left value of the value expression.
     */
    public ValueExp getLeftValue()  {
        return exp1;
    }

    /**
     * Returns the right value of the value expression.
     */
    public ValueExp getRightValue()  {
        return exp2;
    }

    /**
     * Applies the BinaryOpValueExp on a MBean.
     *
     * @param name The name of the MBean on which the BinaryOpValueExp will be applied.
     *
     * @return  The ValueExp.
     *
     * @exception BadStringOperationException
     * @exception BadBinaryOpValueExpException
     * @exception BadAttributeValueExpException
     * @exception InvalidApplicationException
     */
    public ValueExp apply(ObjectName name) throws BadStringOperationException, BadBinaryOpValueExpException,
        BadAttributeValueExpException, InvalidApplicationException  {
        ValueExp val1 = exp1.apply(name);
        ValueExp val2 = exp2.apply(name);
        String sval1;
        String sval2;
        double dval1;
        double dval2;
        long   lval1;
        long   lval2;
        boolean numeric = val1 instanceof NumericValueExp;

        if (numeric) {
            if (((NumericValueExp)val1).isLong()) {
                lval1 = ((NumericValueExp)val1).longValue();
                lval2 = ((NumericValueExp)val2).longValue();

                switch (op) {
                case Query.PLUS:
                    return Query.value(lval1 + lval2);
                case Query.TIMES:
                    return Query.value(lval1 * lval2);
                case Query.MINUS:
                    return Query.value(lval1 - lval2);
                case Query.DIV:
                    return Query.value(lval1 / lval2);
                }

            } else {
                dval1 = ((NumericValueExp)val1).doubleValue();
                dval2 = ((NumericValueExp)val2).doubleValue();

                switch (op) {
                case Query.PLUS:
                    return Query.value(dval1 + dval2);
                case Query.TIMES:
                    return Query.value(dval1 * dval2);
                case Query.MINUS:
                    return Query.value(dval1 - dval2);
                case Query.DIV:
                    return Query.value(dval1 / dval2);
                }
            }
        } else {
            sval1 = ((StringValueExp)val1).getValue();
            sval2 = ((StringValueExp)val2).getValue();

            switch (op) {
            case Query.PLUS:
                return new StringValueExp(sval1 + sval2);
            default:
                throw new BadStringOperationException(opString());
            }
        }

        throw new BadBinaryOpValueExpException(this);
    }

    /**
     * Returns the string representing the object
     */
    public String toString()  {
        try {
            return parens(exp1, true) + " " + opString() + " " + parens(exp2, false);
        } catch (BadBinaryOpValueExpException ex) {
            return "invalid expression";
        }
    }

    /*
     * Add parentheses to the given subexpression if necessary to
     * preserve meaning.  Suppose this BinaryOpValueExp is
     * Query.times(Query.plus(Query.attr("A"), Query.attr("B")), Query.attr("C")).
     * Then the original toString() logic would return A + B * C.
     * We check precedences in order to return (A + B) * C, which is the
     * meaning of the ValueExp.
     *
     * We need to add parentheses if the unparenthesized expression would
     * be parsed as a different ValueExp from the original.
     * We cannot omit parentheses even when mathematically
     * the result would be equivalent, because we do not know whether the
     * numeric values will be integer or floating-point.  Addition and
     * multiplication are associative for integers but not always for
     * floating-point.
     *
     * So the rule is that we omit parentheses if the ValueExp
     * is (A op1 B) op2 C and the precedence of op1 is greater than or
     * equal to that of op2; or if the ValueExp is A op1 (B op2 C) and
     * the precedence of op2 is greater than that of op1.  (There are two
     * precedences: that of * and / is greater than that of + and -.)
     * The case of (A op1 B) op2 (C op3 D) applies each rule in turn.
     *
     * The following examples show the rules in action.  On the left,
     * the original ValueExp.  On the right, the string representation.
     *
     * (A + B) + C     A + B + C
     * (A * B) + C     A * B + C
     * (A + B) * C     (A + B) * C
     * (A * B) * C     A * B * C
     * A + (B + C)     A + (B + C)
     * A + (B * C)     A + B * C
     * A * (B + C)     A * (B + C)
     * A * (B * C)     A * (B * C)
     */
    private String parens(ValueExp subexp, boolean left)
    throws BadBinaryOpValueExpException {
        boolean omit;
        if (subexp instanceof BinaryOpValueExp) {
            int subop = ((BinaryOpValueExp) subexp).op;
            if (left)
                omit = (precedence(subop) >= precedence(op));
            else
                omit = (precedence(subop) > precedence(op));
        } else
            omit = true;

        if (omit)
            return subexp.toString();
        else

/**代码未完, 请加载全部代码(NowJava.com).**/