变量未初始化...但是是吗？

Question

因此，我目前正在处理的应用程序需要3个将值返回到另一个类的方法。 我遇到的问题是第二种方法，该方法将第一种方法的用户输入与相应的“命名月份”相关联。 我不断收到来自此集合的“变量可能尚未初始化”错误：

public String MonthName(int monthNumber){
    /*This method assigns a Name based on the users input and
    **returns the proper name of the corresponding month 
    ** @para: monthNameFin is the final name of the month
    **        rest should be self-explanatory              */

    String monthNameFin;

    if (monthNumber == 1)
        monthNameFin = "January";
    if (monthNumber == 2)
        monthNameFin = "February";
    if (monthNumber == 3)
        monthNameFin = "March";
    if (monthNumber == 4)
        monthNameFin = "April";
    if (monthNumber == 5)
        monthNameFin = "May";
    if (monthNumber == 6)
        monthNameFin = "June";
    if (monthNumber == 7)
        monthNameFin = "July";
    if (monthNumber == 8)
        monthNameFin = "August";
    if (monthNumber == 9)
        monthNameFin = "September";
    if (monthNumber == 10)
        monthNameFin = "October";
    if (monthNumber == 11)
        monthNameFin = "November";
    if (monthNumber == 12)
        monthNameFin = "December";

    return monthNameFin;
}

任何帮助是极大的赞赏！

Answer 1

您必须将变量初始化为：

String monthNameFin=null;

要么

String monthNameFin="";

否则，如果您始终有一个值分配，则可以跳过初始化，例如：

String monthNameFin;
   if (monthNumber == 1){
        monthNameFin = "January";
   }else{ 
        monthNameFin = "January";
   }

在这种情况下，无法逃脱……到monthNumberFin无论如何都将是一个值，而在您的代码中，它可能永远不会输入if ..假设它是例如montthumber = 14

Answer 2

如果monthNumber小于或等于0，或者大于等于13，则总是有可能没有匹配的案例。 这意味着存在变量未初始化的可能性。

由于超出1-12范围的数字没有意义，因此如果没有大小写匹配，最好在底部抛出IllegalArgumentException 。

此示例还包括将所有案例分配语句更改为return语句，以使每个案例（即使是不匹配的案例）都有return或throw 。

  ...
  if (monthNumber == 12)
      return "December";

  // No case matched.
  throw new IllegalArgumentException("Bad month number: " + monthNumber);
}

调用此方法的代码应捕获此异常并正确处理它。

Answer 3

正如其他答案所述，您必须初始化变量。

但是， if不是所有这些，我将使用DateFormatSymbols对象。

public String MonthName(int monthNumber){
     if(monthNumber < 1 || monthNumber > 12)
           throw new IllegalArgumentException("Month must be in the range [1, 12]");
     DateFormatSymbols dfs = DateFormatSymbols.getInstance(Locale.UK);
     return dfs.getMonths()[monthNumber-1];
}

Answer 4

可以处理所有整数情况的更简单方法：

public String monthName(int monthNumber) {
    if (monthNumber < 1 || monthNumber > 12) return null;
    String[] months = new String[] {"January", "February", "March", "April", "May",
            "June", "July", "August", "September", "October", "November", "December"};
    return months[monthNumber];
}

也，

（1）首先只初始化monthName，也许是

String month = monthName(monthNumber);

（2）您应该以小写字母开头方法名称。

Answer 5

不应在输入语句“ IF”中找到未初始化的变量。

初始化monthNameFin。

尝试这个：

public String MonthName(int monthNumber){
    /*This method assigns a Name based on the users input and
    **returns the proper name of the corresponding month 
    ** @para: monthNameFin is the final name of the month
    **        rest should be self-explanatory              */

    String monthNameFin = ""; /* <----------- */

    if (monthNumber == 1)
        monthNameFin = "January";
    if (monthNumber == 2)
        monthNameFin = "February";
    if (monthNumber == 3)
        monthNameFin = "March";
    if (monthNumber == 4)
        monthNameFin = "April";
    if (monthNumber == 5)
        monthNameFin = "May";
    if (monthNumber == 6)
        monthNameFin = "June";
    if (monthNumber == 7)
        monthNameFin = "July";
    if (monthNumber == 8)
        monthNameFin = "August";
    if (monthNumber == 9)
        monthNameFin = "September";
    if (monthNumber == 10)
        monthNameFin = "October";
    if (monthNumber == 11)
        monthNameFin = "November";
    if (monthNumber == 12)
        monthNameFin = "December";

    return monthNameFin;
}