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JDK1.4, 1.5 的 String Class 代码如下 [code] public final class String implements java.io.Serializable, Comparable<String>, CharSequence { /** The value is used for character storage. */ private final char value[]; /** The offset is the first index of the storage that is used. */ private final int offset; /** The count is the number of characters in the String. */ private final int count; [/code] [code] /** * Initializes a newly created <code>String</code> object so that it * represents the same sequence of characters as the argument; in other * words, the newly created string is a copy of the argument string. Unless * an explicit copy of <code>original</code> is needed, use of this * constructor is unnecessary since Strings are immutable. * * @param original a <code>String</code>. */ public String(String original) { int size = original.count; char[] originalValue = original.value; char[] v; if (originalValue.length > size) { // The array representing the String is bigger than the new // String itself. Perhaps this constructor is being called // in order to trim the baggage, so make a copy of the array. v = new char[size]; System.arraycopy(originalValue, original.offset, v, 0, size); } else { // The array representing the String is the same // size as the String, so no point in making a copy. v = originalValue; } this.offset = 0; this.count = size; this.value = v; } [/code] 从这段构造函数中，我们可以看出，不同Reference的String之间有可能共享一样的 char[]。 [code] /** * Compares this string to the specified object. * The result is <code>true</code> if and only if the argument is not * <code>null</code> and is a <code>String</code> object that represents * the same sequence of characters as this object. * * @param anObject the object to compare this <code>String</code> * against. * @return <code>true</code> if the <code>String </code>are equal; * <code>false</code> otherwise. * @see java.lang.String#compareTo(java.lang.String) * @see java.lang.String#equalsIgnoreCase(java.lang.String) */ public boolean equals(Object anObject) { if (this == anObject) { return true; } if (anObject instanceof String) { String anotherString = (String)anObject; int n = count; if (n == anotherString.count) { char v1[] = value; char v2[] = anotherString.value; int i = offset; int j = anotherString.offset; while (n-- != 0) { if (v1[i++] != v2[j++]) return false; } return true; } } return false; } [/code] 但是，equals 方式好像忽略了这个可能。没有直接对两者的char[]的reference进行比较。 按照我的想法，应该加入这么一段。 [code] if (anObject instanceof String) { String anotherString = (String)anObject; int n = count; if (n == anotherString.count) { char v1[] = value; char v2[] = anotherString.value; int i = offset; int j = anotherString.offset; ////{{ if(i == j && v1 == v2) return true; // NOTE: this line is added by me ////}} while (n-- != 0) { if (v1[i++] != v2[j++]) return false; } [/code] 这样就能够对应共享 char[] 的情况，能够加快比较速度。 返回类别: 教程 上一教程: Java的double类型探索. 下一教程: Java多线程程序设计 您可以阅读与"Java String 的 equals() 方式可能的优化"相关的教程: · Java性能的优化（上） · (代码级)Java性能的优化 · Java代码优化－－尽可能地使用stack(栈)变量（方式内部的局部变量） · Java程序的性能优化StringBuffer与Vector · Java 性能优化技巧集锦 （2） |
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