标签:param static 数字 ++ 结果 new int [] arp
输出结果
5 4 3 2 1 a
4 3 2 1 b
3 2 1 c
2 1 d
1 e
public class Test { /** * @param args */ public static void main(String[] args) { NumPrinter numPrinter = new NumPrinter(); CharPrinter charPrinter = new CharPrinter(); int charst = 97; for (int st = 5; st >= 1; st--) { numPrinter.print(st); charPrinter.print(charst++); System.out.println(); } } } class NumPrinter { public void print(int i) { for (; i >= 1; i--) { System.out.print(i + " "); } } } class CharPrinter { public void print(int i) { System.out.print((char) i); } }
关键是字符的输出,因为前面的数字是递减的,而字符是递增的,因此需要使用数字强转成char
标签:param static 数字 ++ 结果 new int [] arp
原文地址:https://www.cnblogs.com/yszzu/p/9162606.html