摘要： 原因：JAVA核心编程1卷7版P116 Employee E1 = new Employee("Alice",...); Employee E2 = new Employee("Bob",...); swap(E1, E2);void swap(Employee a, Employee b){ Employee temp; temp = a; a = b; b = temp;}无法实现，E1， E2 的引用互换！！！故是按值传递的。 阅读全文