Question

人们..我已经查过类似的问题，但没有找到与我类似的东西。在行

中抛出异常后，我得到“List无法解析为类型”

else
    throw new ExceptionLogicMalformedSignal(c, "Invalid character!");

在fromString(char c)方法中。我在代码中做错了什么？为什么会发生非列表方法？提前谢谢！

import java.util.LinkedList;
import java.util.List;


public enum Signal {

    HI, LO, X;


    public Signal invert()
    {
        if(this == HI)
            return LO;
        else if(this == LO)
            return HI;
        else if(this == X)
            return X;

        return this;
    }

    public static Signal fromString(char c)
    {

        if(c == '1')
            return HI;
        else if(c == '0')
            return LO;
        else if(c == 'X')
            return X;
        else if(c == 'x')
            return X;
        else
            throw new ExceptionLogicMalformedSignal(c, "Invalid character!");

    }

    public List <Signal> fromString(String inps)
    {
        List<Signal> values = new LinkedList<Signal>();
        if(inps.equals("1"))
           values.add(HI);
        else if(inps.equals("0"))
            values.add(LO);
        else if(inps.equals("X"))
            values.add(X);
        else if(inps.equals("x"))
            values.add(X);
        else 
            throw new ExceptionLogicMalformedSignal(inps.charAt(0), "Invalid character!");

        return values;

    }

    @Override
    public String toString()
    {
        if(this == HI)
            return "1";
        else if(this == LO)
            return "0";
        else if(this == X)
            return "X";

        return "Error here!";
        }
    }



public class ExceptionLogicMalformedSignal extends RuntimeException {

    private char bad; //offending character 
    private String msg = "You entered an invalid character"; //the entire message you want to convey

    public ExceptionLogicMalformedSignal(char bad, String msg)
    {
        this.bad = bad;
        this.msg = msg;
    }

    @Override
    public String toString()
    {
        return msg;
    }

    public String getBad()
    {
        String value = Character.toString(this.bad);
        return value;
    }

    public String msg()
    {
        return msg;
    }

    public void setBad(String bad)
    {
        String value = Character.toString(this.bad);
        value = bad;
    }

    public void setMsg(String msg)
    {
        this.msg = msg;
    }

}

Answer 1

List cannot be resolved to a type

是一个编译器错误。在抛出异常之后你就得不到它了＃39;在编译使用List.

的代码行时，您会得到它

你需要

import java.util.List;

尽管你发布了什么，但显然它不存在。

列表无法解析为类型

1 个答案: