Unreachable statement?(无法到达的声明?)
问题描述
这是我的编码:出于某种原因,最后,当我试图返回 winscounter 和 losscounter 时,它说的是无法访问的语句,但不是针对 tiescounter 的.我不知道为什么!如果有人能回答这个问题,将不胜感激!
Here is my coding: For some reason, at the very end, when I'm trying to return winscounter and losscounter, it says unreachable statement but not for the tiescounter. I can't figure out why! If anyone can answer this, it would be greatly appreciated!!
public class RockPaperScissors {
/**
* @param args the command line arguments
*/
static int value; //computer's choice
static int choice; //user choice
static int tiescounter = 0;
static int winscounter = 0;
static int losscounter = 0;
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader (new InputStreamReader (System.in));// user input
int repeat;
do {
System.out.println("ROCK PAPER SCISSORS"+
"
===================");
System.out.println("
1=Rock" +
"
2=Paper" +
"
3=Scissors" +
"
===========" +
"
Choose:");
choice = Integer.parseInt(br.readLine());
if (choice !=1 && choice !=2 && choice !=3) {
do{
System.out.println("
Error. Please choose Rock, Paper or Scissors.");
choice = Integer.parseInt(br.readLine());
}
while (choice !=1 && choice !=2 && choice !=3);
}
System.out.println();
if (choice == 1){
System.out.println("You have chosen Rock.");
}
else if (choice ==2){
System.out.println("You have chosen Paper.");
}
else if(choice == 3){
System.out.println("You have chosen Scissors.");
}
randomWholeNumber();
if (value == 1){
System.out.println("The computer has chosen Rock." );
}
else if (value == 2){
System.out.println("The computer has chosen Paper." );
}
else if (value == 3){
System.out.println("The computer has chosen Scissors." );
}
determineOutcome();
System.out.println("Ties:"+ tiescounter);
System.out.println("Wins: " + winscounter);
System.out.println("Losses: " + losscounter);
repeat = Integer.parseInt(br.readLine());
}
while (repeat==1);
}
public static int randomWholeNumber(){
do{
value=0;//resets random number
//generates and returns a random number within user's range
value = (int) ((Math.random()*3)+1);
}
while((value>3)||(value<1));
return (value);
}
public static int determineOutcome(){
if (value == choice){
System.out.println("
YOU'VE TIED");
do{
tiescounter+=1;
}
while (tiescounter != tiescounter);
}
else if (value == 1){ //Rock
if (choice == 2){ //Paper
System.out.println("
YOU'VE WON");
do{
winscounter +=1;
}
while (winscounter != winscounter);
}
else if (choice == 3){ //Scissors
System.out.println("
YOU'VE LOST");
do{
losscounter+=1;
}
while(losscounter!=losscounter);
}
}
else if (value == 2){ //Paper
if (choice == 1){ //Rock
System.out.println("
YOU'VE LOST");
do{
losscounter+=1;
}
while(losscounter!=losscounter);
}
else if (choice == 3){ //Scissors
System.out.println("
YOU'VE WON");
do{
winscounter +=1;
}
while (winscounter != winscounter);
}
}
else if (value == 3){ //Scissors
if (choice == 1){ //Rock
System.out.println("
YOU'VE WON");
do{
winscounter +=1;
}
while (winscounter != winscounter);
}
else if (choice == 2){ //Paper
System.out.println("
YOU'VE LOST");
do{
losscounter+=1;
}
while(losscounter!=losscounter);
}
}
return(tiescounter);
return(winscounter);
return(losscounter);
}
}
推荐答案
一个函数只能返回一个值.一旦 return(tiescounter);
被执行,函数就会退出.如果要返回所有三个值,则必须将它们包装在一个类中.
A function can only return one value. As soon as return(tiescounter);
is executed the function exits. If you want to return all three values you will have to wrap them in a class.
顺便说一下,return(tiescounter);
可以写成 return tiescounter;
不需要在返回值周围加上括号.两个语句的结果相同.
By the way return(tiescounter);
can be written as return tiescounter;
Parenthesis around return values is not required. Both statements will have the same result.
这篇关于无法到达的声明?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!
本文标题为:无法到达的声明?
基础教程推荐
- 减少 JVM 暂停时间 >1 秒使用 UseConcMarkSweepGC 2022-01-01
- 降序排序:Java Map 2022-01-01
- 在 Libgdx 中处理屏幕的正确方法 2022-01-01
- 如何使用 Java 创建 X509 证书? 2022-01-01
- Java:带有char数组的println给出乱码 2022-01-01
- 设置 bean 时出现 Nullpointerexception 2022-01-01
- Java Keytool 导入证书后出错,"keytool error: java.io.FileNotFoundException &拒绝访问" 2022-01-01
- “未找到匹配项"使用 matcher 的 group 方法时 2022-01-01
- FirebaseListAdapter 不推送聊天应用程序的单个项目 - Firebase-Ui 3.1 2022-01-01
- 无法使用修饰符“public final"访问 java.util.Ha 2022-01-01