方法的重载（overLoad）

public class overLoad {
	public static void main(String[] args) {
		Metool metool=new Metool();
		metool.m(10);
		metool.m("hallow");
		metool.m(10, 30);
	}
}
class Metool{
	public void m(int n){
		System.out.println("m的平方="+(n*n));
	}
	public void m(int n1,int n2){
		System.out.println("m的积="+(n1*n2));
	}
	public void m(String str){
		System.out.println("传入的值是："+str);
	}
}

		int aint =metool.max(10, 30);
		System.out.println(aint);
		System.out.println(metool.max(1.1, 23.1));
		System.out.println(metool.max(31.5, 15.1, 1));
		
	}
}
class Metool{
	public int max(int n1,int n2){
		return n1>n2 ? n1:n2;
	}
	public double max(double n1,double n2){
		return n1>n2 ? n1:n2;
	}
	public double max(double n1,double n2,double n3){
		double max1= n1>n2  ? n1:n2;
		return max1>n3 ? max1:n3;
	}