针对java中String源码hashcode算法源码分析
/** The value is used for character storage. */ private final char value[]; //将字符串截成的字符数组 /** Cache the hash code for the string */ private int hash; // Default to 0 用以缓存计算出的hashcode值 /** * 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; }
按照上面源码举例说明:
String msg = "abcd"; // 此时value[] = {'a','b','c','d'} 因此
for循环会执行4次
第一次:h = 31*0 + a = 97
第二次:h = 31*97 + b = 3105
第三次:h = 31*3105 + c = 96354
第四次:h = 31*96354 + d = 2987074
由以上代码计算可以算出 msg 的hashcode = 2987074 刚好与 System.err.println(new String("abcd").hashCode()); 进行验证
在源码的hashcode的注释中还提供了一个多项式计算方式:
s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
s[0] :表示字符串中指定下标的字符
n:表示字符串中字符长度
a*31^3 + b*31^2 + c*31^1 + d = 2987074 + 94178 + 3069 + 100 = 2987074 ;