Java 字符串变量可能尚未初始化

String Variable might not have been initialized

提问by Boxasauras

Here is my code:

这是我的代码：

import java.util.Scanner;

public class empMod

{
public static void main(String[] args)
    {
    int choice;

    Scanner input = new Scanner(System.in);

    do
        {
        choice = -1;
        System.out.println("Employee Data:");
        System.out.println("1. - Employee Name:");
        System.out.println("2. - Employee Hire Date:");
        System.out.println("3. - Employee Address:");
        System.out.println("4. - Employee Number:");
        System.out.println("5. - Exit");

        choice = input.nextInt();
        input.nextLine();

        switch (choice)
            {
            case 1:

            String empName = new String ();
            System.out.println("Enter the name of the employee:");
            String name = input.nextLine();
            break;

            case 2:

            String empDate = new String ();
            System.out.println("Enter the hire date of the employee:");
            String date = input.nextLine();
            break;

            case 3:

            String empAddress = new String ();
            System.out.println("Enter the address of the employee:");
            String address = input.nextLine();
            break;

            case 4:

            String empNumb = new String ();
            System.out.println("Enter the Employee number:");
            int number = input.nextInt();
            break;

            case 5:

            System.out.print("\n");
            System.out.println("The name of the employee is: " + empName); // <-- This is the line where the error occurs.
            break;

            default:
            continue;
            }

        }
    while (choice != 6);
    } 
}

The intent of the program is to have the user input information about the employee, and then at request, have the information displayed. When I go to compile the program, I get the following error:

该程序的目的是让用户输入有关员工的信息，然后根据要求显示该信息。当我去编译程序时，出现以下错误：

empMod.java:57: error: variable empName might not have been initialized
                                System.out.println("The name of the employee is:
 " + empName);

     ^

The string variable is initialized in another case though, so I am not sure of the problem.

不过，字符串变量在另一种情况下被初始化，所以我不确定这个问题。

回答by dantuch

switch (choice)and cases later means, that according to value in switch(choice) one of casewill be choosen. If it will be 5, your variable will not be initialized.

switch (choice)和cases 后面的意思是，根据switch（选择）中的值，case将选择其中之一。如果是5，您的变量将不会被初始化。

you need to initialize empNamebefore switch, of in every casein whitch it's used.

您需要在每次使用empName之前进行初始化。switchcase

And you should not use String empName = new String ();but String empName = "";- it will use String Pool.

你不应该使用String empName = new String ();但是String empName = "";- 它会使用字符串池。

回答by JB Nizet

The empName variable is only initialized in the case 1section. So what would happen if this block was never executed, and the case 5section was? What would be printed, since the variable has never been initialized to anything?

empName 变量仅在case 1节中初始化。那么如果这个块从来没有被执行过，而这个case 5部分却被执行了，会发生什么？将打印什么，因为变量从未被初始化为任何东西？

Add

添加

String empName = "";

before the loop.

在循环之前。

回答by Sotirios Delimanolis

A switchblock contains a scope. Variables declared in that scope can only be used there. Think of the switchas multiple if-elses. If the if(the case) that initializes you variable doesn't get executed, then you are left with an uninitialized variable. That's what the compiler is complaining about.

一个switch块包含一个范围。在该范围内声明的变量只能在该范围内使用。将switch视为多个if-elses。如果初始化变量的if(the case) 没有被执行，那么你就会得到一个未初始化的变量。这就是编译器所抱怨的。

In

在

case 5:
    System.out.print("\n");
    System.out.println("The name of the employee is: " + empName); // <-- This is the line where the error occurs.
    break;

empNameis only initialized if execution flowed through all the cases (ie. choicehad value 1 and your cases did not have breaks). But the compiler cannot be sure of this.

empName仅在执行流经所有案例时才初始化（即choice具有值 1 并且您的案例没有中断）。但是编译器不能确定这一点。

The way to fix this is to declare your empNamevariable outside the switchblock so that its scope is method scope and not limited to inside the switch. You would need to initialize it to some default value so that the compiler knows that it is initialized.

解决这个问题的方法是empName在switch块外声明你的变量，这样它的范围就是方法范围而不限于switch. 您需要将其初始化为某个默认值，以便编译器知道它已初始化。

String empName = "Not a Name";

回答by GoingMyWay

Because int case 5, you didn't define String empName = new String ();, you can define empName in main

因为 int case 5，你没有定义String empName = new String ();，你可以在 main 中定义 empName

public static void main(String[] args)
{
    int choice;
    String empName = "Not a name";
    Scanner input = new Scanner(System.in);

...