重载的好处
- 减轻了起名的麻烦
- 减轻了记名的麻烦
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
| public class OverLoad01 {
public static void main(String[] args) { MyCalculator mc = new MyCalculator(); System.out.println(mc.calculate(1, 2)); System.out.println(mc.calculate(1.1, 2)); System.out.println(mc.calculate(1, 2.1)); } } class MyCalculator {
public int calculate(int n1, int n2) { System.out.println("calculate(int n1, int n2) 被调用"); return n1 + n2; } public double calculate(int n1, double n2) { return n1 + n2; } public double calculate(double n1, int n2) { System.out.println("calculate(double n1, int n2) 被调用.."); return n1 + n2; } public int calculate(int n1, int n2,int n3) { return n1 + n2 + n2; } }
|
注意事项以及使用细节
- 方法名:必须相同
- 形参列表:必须不同(形参类型或个数或顺序,至少有一样不同,参数名无要求)
- 返回类型:无要求
练习
1.判断题:
与voidshow(int a,char b,doublec){}构成重载的有:[2 3 4 5]
1 2 3 4 5 6 7
| 1. void show(int x,char y,double z){} 2. int show(int a,double c,char b){} 3. void show(int a ,double c,char b){} 4. boolean show(int c,char b){} 5. void show(double c){} 6. double show(int x,char y,double z){} 7. voidshows(){}
|