纵观java源代码,我在类AbstractList的hashCode()方法中遇到了难以理解的构造。这是用于hashCode的ArrayList方法的实现。我不明白它是如何迭代的。public int hashCode() { for (E e : this) //<--???hashCode = 31*hashCode + (e==null ? 0 : e.hashCode());
当我查看Android Pair的hashCode的“旧”版本时,它的实现如下所示 * Compute a hash code usingthe hash codes of the underlying objects */ int result = 17;
result = 31 * result + first.<em
我需要为teams类编写一个equals()方法,该方法与提供的hashcode方法一致 public int hashCode() return this.getPro().hashCode() }public boolean equals(Object obj) ClassName pro = (ClassN
public static int getHash(String str, int limit) int hashCode = Math.abs(str.hashCode()%(limit));}
* @param argspublic static void main(String[] args) int hashLimit= getHash(curStr,hashLimit);
if ( arr[hash
Object.hashCode()的默认行为本质上是返回对象的“地址”,以便a.hashCode() == b.hashCode()当且仅当是== b.如果超类已经定义了hashCode(),我如何在用户定义的类中获得这种行为例如: public int hashCode() { } public int hashCode() {
// Now I wan