Java equals 和 hashcode 方法
问题
面试时经常会问起字符串比较相关的问题,
总结一下,大体是如下几个:
1.字符串比较时用的什么方法,内部实现如何?
2.hashcode的作用,以及重写equal方法,为什么要重写hashcode方法?
现在对以上几个问题,彻底做一个研究和小结.
PS:本文使用jdk1.7
解析
1.Object类 的equals 方法
/** * 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
1.reflexive-自反性
x.equals(x) return true
2.symmetric-对称性
x.equals(y) return true
y.equals(x) return true
3.transitive-传递性
x.equals(y) return true
y.equals(z) return true
x.equals(z) return true
4.consistent-一致性
x.equals(y) return true //那么不管调用多少次,肯定都是返回true
5.与null的比较
x.equals(null) return false //对于none-null的x对象,每次必然返回false
6.于hashcode的关系
* 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
2.String类的equals方法
/** * 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; }
看源码,我们可以发现,这个比较分为两部分
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
小结
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
1.字符串比较时用的什么方法,内部实现如何?
2.hashcode的作用,以及重写equal方法,为什么要重写hashcode方法?