网上找到特么一段话:
Java对于equals方法和hashCode方法是这样规定的: 1、如果两个对象相等,那么它们的hashCode值一定要相等; 2、如果两个对象的hashCode相等,它们并不一定相等。PS:相等说的是equals方法。
那么这2个方法是什么来的??
先祭出源码。
equals源码:
/** * 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 = 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; }
equals方法, 关键点就是把字符串当成字符数组(char[]),然后两个相同长度的字符数组每个相同下标的字符进行比较。
hashCode源码:
/** * 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; int len = count; if (h == 0 && len > 0) { int off = offset; char val[] = value; for (int i = 0; i < len; i++) { h = 31*h + val[off++]; } hash = h; } return h; }
hashCode是对字符串每一个字符做一个简单“编码”后,累加值。并不严格规定每一个字符相等。
时间: 2024-10-19 00:55:40