Java 编译器给我一个';无法到达的状态';错误
这是我代码中的一个方法,当我试图编译它时,它会抛出一个“无法访问的语句”错误Java 编译器给我一个';无法到达的状态';错误,java,compiler-errors,Java,Compiler Errors,这是我代码中的一个方法,当我试图编译它时,它会抛出一个“无法访问的语句”错误 public static boolean whoareyou(String player) { boolean playerwhat; if (player.equalsIgnoreCase("Player 1")) { return true; } else { return false; } return player
public static boolean whoareyou(String player)
{
boolean playerwhat;
if (player.equalsIgnoreCase("Player 1"))
{
return true;
}
else
{
return false;
}
return playerwhat;
}
确切的错误是:
java:82: error: unreachable statement
return playerwhat;
^
然后,我尝试使用以下代码中返回的布尔值:
public static int questions(int diceroll, int[] scorep1)
{
String wanttocont = " ";
boolean playerwhat;
for (int i = 0; i <= 6; i++)
{
while (!wanttocont.equalsIgnoreCase("No"))
{
wanttocont = input("Do you wish to continue?");
// boolean playerwhat; wasn't sure to declare here or outside loop
if (diceroll == 1)
{
String textinput = input("What's 9+10?");
int ans1 = Integer.parseInt(textinput);
output("That's certainly an interesting answer.");
if (ans1 == 19)
{
if (playerwhat = true)
{
output("Fantastic answer player 1, that's correct!");
diceroll = dicethrow(diceroll);
scorep1[0] = scorep1[0] + diceroll;
output("Move forward " + diceroll + " squares. You are on square " + scorep1[0]);
}
else if (playerwhat = false)
{
output("Fantastic answer player 2, that's correct!");
diceroll = dicethrow(diceroll);
scorep1[1] = scorep1[1] + diceroll;
output("Move forward " + diceroll + " squares. You are on square " + scorep1[1]);
}
} // END if diceroll is 1
} // END while wanttocont
} // END for loop
} // END questions
公共静态整数问题(整数掷骰子,整数[]分数P1)
{
字符串wanttocont=“”;
布尔playerwhat;
for(int i=0;ireturn playerwhat;
永远无法到达,因为if
或else
子句将返回true
或false
。因此您应该删除此语句。playerwhat
变量不是必需的
顺便说一句,您的方法可以替换为一行方法:
public static boolean whoareyou(String player)
{
return player.equalsIgnoreCase("Player 1");
}
我会将此方法重命名为更具描述性的方法,例如isFirstPlayer
编辑:
你从不打电话给你是谁
是你的问题
方法。你应该叫它:
替换
if (playerwhat = true) // this is assigning true to that variable, not comparing it to true
与
只需以这种方式更新代码
public static boolean whoareyou(String player)
{
boolean playerwhat;
if (player.equalsIgnoreCase("Player 1"))
{
playerwhat = true;
}
else
{
playerwhat = false;
}
return playerwhat;
}
试试这个:
public static boolean whoareyou(String player)
{
return player.equalsIgnoreCase("Player 1");
}
你有这个问题,因为:
从未到达。您可以通过“if”-或“else”退出函数-部分。你是说要删除返回播放器什么;
在你是谁
方法的末尾?如果我这样做,那么就没有任何东西被传递到问题
方法,就像我在回答了一个问题后说我是玩家2,它仍然会说奇妙的回答玩家1,这是正确的!
当它发出声音时我真的会说player 2
@Overclock请参见编辑。这是你的问题方法的另一个问题。非常感谢!我对你说的话做了一些调整,但效果确实很好。我更喜欢if(“player 1”。equalsIgnoreCase(player))
@SpringLearner同意我们也可以这样做,但我只是按照他的原始代码发布
public static boolean whoareyou(String player)
{
return player.equalsIgnoreCase("Player 1");
}
return player what;