[b]PS:本文使用jdk1.7
[/b]解析
[b]
1.Object类 的equals 方法
[/b]
/**
* Indicates whether some other object is "equal to" this one.
* <p>
* The {@code equals} method implements an equivalence relation
* on non-null object references:
* <ul>
* <li>It is <i>reflexive</i>: for any non-null reference value
* {@code x}, {@code x.equals(x)} should return
* {@code true}.
* <li>It is <i>symmetric</i>: for any non-null reference values
* {@code x} and {@code y}, {@code x.equals(y)}
* should return {@code true} if and only if
* {@code y.equals(x)} returns {@code true}.
* <li>It is <i>transitive</i>: for any non-null reference values
* {@code x}, {@code y}, and {@code z}, if
* {@code x.equals(y)} returns {@code true} and
* {@code y.equals(z)} returns {@code true}, then
* {@code x.equals(z)} should return {@code true}.
* <li>It is <i>consistent</i>: for any non-null reference values
* {@code x} and {@code y}, multiple invocations of
* {@code x.equals(y)} consistently return {@code true}
* or consistently return {@code false}, provided no
* information used in {@code equals} comparisons on the
* objects is modified.
* <li>For any non-null reference value {@code x},
* {@code x.equals(null)} should return {@code false}.
* </ul>
* <p>
* The {@code equals} method for class {@code Object} implements
* the most discriminating possible equivalence relation on objects;
* that is, for any non-null reference values {@code x} and
* {@code y}, this method returns {@code true} if and only
* if {@code x} and {@code y} refer to the same object
* ({@code x == y} has the value {@code true}).
* <p>
* Note that it is generally necessary to override the {@code hashCode}
* method whenever this method is overridden, so as to maintain the
* general contract for the {@code hashCode} method, which states
* that equal objects must have equal hash codes.
*
* @param obj the reference object with which to compare.
* @return {@code true} if this object is the same as the obj
* argument; {@code false} otherwise.
* @see #hashCode()
* @see java.util.HashMap
*/
public boolean equals(Object obj) {
return (this == obj);
}
看代码,Object的equals方法,采用== 进行比较,只是比较对象的引用,如果引用的对象相同,那么就返回true.
看注释,Object的equals方法,具有如下特性
[b]1.reflexive-自反性
[/b] x.equals(x) return true
[b]2.symmetric-对称性
[/b]x.equals(y) return true
y.equals(x) return true
[b]3.transitive-传递性
[/b]x.equals(y) return true
y.equals(z) return true
x.equals(z) return true
[b]4.consistent-一致性
[/b]x.equals(y) return true //那么不管调用多少次,肯定都是返回true
[b]5.与null的比较
[/b]x.equals(null) return false //对于none-null的x对象,每次必然返回false
[b]6.于hashcode的关系
[/b] * Note that it is generally necessary to override the {@code hashCode}
* method whenever this method is overridden, so as to maintain the
* general contract for the {@code hashCode} method, which states
* that equal objects must have equal hash codes.
需要注意的是,一般来说,如果重写了equals方法,都必须要重写hashcode方法,
来确保具有相同引用的对象,能够具有同样的hashcode值
好了,看到这里,我们就明白了,为什么重写了equals方法,一般来说就需要重写hashcode方法,
虽然这个不是强制性的,但是如果不能保证相同的引用对象,没有相同的hashcode,会对系统留下很大隐患
[b]
2.String类的equals方法
[/b]
/**
* Compares this string to the specified object. The result is {@code
* true} if and only if the argument is not {@code null} and is a {@code
* String} object that represents the same sequence of characters as this
* object.
*
* @param anObject
* The object to compare this {@code String} against
*
* @return {@code true} if the given object represents a {@code String}
* equivalent to this string, {@code false} otherwise
*
* @see #compareTo(String)
* @see #equalsIgnoreCase(String)
*/
public boolean equals(Object anObject) {
if (this == anObject) {
return true;
}
if (anObject instanceof String) {
String anotherString = (String) anObject;
int n = value.length;
if (n == anotherString.value.length) {
char v1[] = value;
char v2[] = anotherString.value;
int i = 0;
while (n-- != 0) {
if (v1[i] != v2[i])
return false;
i++;
}
return true;
}
}
return false;
}
[b]看源码,我们可以发现,这个比较分为两部分
[/b]1.先比较是否引用同一对象
2.如果引用对象不同,是否两个String的content相同
3,String 类的hashcode 方法
/**
* Returns a hash code for this string. The hash code for a
* <code>String</code> object is computed as
* <blockquote><pre>
* s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
* </pre></blockquote>
* using <code>int</code> arithmetic, where <code>s[i]</code> is the
* <i>i</i>th character of the string, <code>n</code> is the length of
* the string, and <code>^</code> indicates exponentiation.
* (The hash value of the empty string is zero.)
*
* @return a hash code value for this object.
*/
public int hashCode() {
int h = hash;
if (h == 0 && value.length > 0) {
char val[] = value;
for (int i = 0; i < value.length; i++) {
h = 31 * h + val[i];
}
hash = h;
}
return h;
}
可以看到hashcode的计算公式为:s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
因此,对于同一个String,得出的hashcode必然是一致的
另外,对于空的字符串,hashcode的值是0
[b]
小结
[/b]至此,我们可以对本文开头的疑问做一个小结.
1.字符串比较时用的什么方法,内部实现如何?
使用equals方法,先比较引用是否相同,后比较内容是否一致.
2.hashcode的作用,以及重写equal方法,为什么要重写hashcode方法?
hashcode是系统用来快速检索对象而使用,equals方法是用来判断引用的对象是否一致,所以,当引用对象一致时,必须要确保其hashcode也一致,因此需要重写hashcode方法来确保这个一致性
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