用-Xlint重新编译：未经检查以获取详细信息 [英] Recompile with -Xlint : unchecked for details

问题描述

我的代码中有此警告。我尝试使用

Hi guys I have this warning on my code. I try to compile using

@SuppressWarnings（ unchecked）

@SuppressWarnings("unchecked")

进行编译

它可以编译，但是在执行时程序找不到主类。

It compile but at the time of the execution the program can't find the main class.

如何解决此问题？

下面的代码：

import java.util.*;

@SuppressWarnings("unchecked")

class ProgRub {
    public static void main(String argv[]) {
        Hashtable rubrica = new Hashtable(20);
        String chiave, valore;
        Menu mioMenu = new Menu();
        int scelta;
        scelta = (int) mioMenu.scelta();

        while (scelta != 5) {
            if (scelta == 1) {
                chiave = mioMenu.leggiDato("Nome:");
                valoe = mioMenu.leggiDato("Valore:");
                rubrica.put(chiave, valore);
            } else if (scelta == 2) {
                chiave = mioMenu.leggiDato("Nome:");
                rubrica.remove(chiave);
            } else if (scelta == 3) {
                Iterator i = rubrica.keySet().iterator();
                while (i.hasNext()) {
                    chiave = (String) i.next();
                    valore = (String) rubrica.get(chiave);
                    System.out.println(chiave + "tel." + valore);
                }
            }

            else if (scelta == 4) {
                chiave = mioMenu.leggiDato("Nome:");
                if (rubrica.contains(chiave)) {
                    valore = (String) rubrica.get(chiave);
                    System.out.println("Telefono:" + valore);
                }

                else {
                    System.out.println("Nominativo inesistente.");
                }
            }
            scelta = (int) mioMenu.scelta();
        }

        System.out.println("Fine programma.");
    }
}

推荐答案

您使用 Hashtable 和 Iterator 作为原始类型。如果您分别将它们用作 Hashtable< String，String> 和 Iterator< String> ：

You are using Hashtable and Iterator as raw types. If you use them as Hashtable<String, String> and Iterator<String> respectively:

• 您不必抑制未经检查的警告，并且


• 您可以摆脱一堆类型转换

要了解更多信息，建议您花些时间阅读Java泛型：

To understand more, I recommend that you take the time to read up on Java generics:

• https： //docs.oracle.com/javase/tutorial/java/generics/index.html

• https://docs.oracle.com/javase/tutorial/java/generics/index.html

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