Question

我尝试捕捉以防止输入不匹配。我是java的新手，可能缺少对方法的一些理解。每当我把try catch它给我一个关于必须创建一个返回类型的错误，即使我已经在finally块中。如果我把它保留在try块中，它说我需要一个返回类型。继承我的代码：

      public static double average() {
       try{
    Scanner avgs = new Scanner(System.in);
    System.out.println("Enter total number of numbers; ");
    double tnum = avgs.nextDouble();
    double[] nums = new double[(int) tnum];
    double sum = 0;
    for (int i = 0; i < tnum; i++) {
        System.out.println("Enter number " + (i + 1) + " : ");
        nums[i] = avgs.nextDouble();
        sum += nums[i];

    }

    System.out.println(" ");
    System.out.println("The sum is: " + sum);
    double avg = sum / tnum;
    System.out.println("The average is: " + avg);



}catch(InputMismatchException e){System.out.println("Enter a number honey:);                     

}
   finally{return avg;}

}

错误：

  Enter a for average or s for  sum
  a
  Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
avg cannot be resolved to a variable

at com.towerdef.shit.Recursion.average(Recursion.java:78)
at com.towerdef.shit.Recursion.yn(Recursion.java:18)
at com.towerdef.shit.Recursion.main(Recursion.java:31)

Answer 1

应该是：

public static double average() {
   double avg;
   try{

并进一步向下：

System.out.println("The sum is: " + sum);
avg = sum / tnum;
System.out.println("The average is: " + avg);

那就行了。正如其他人所说，avg的定义方式只是包含它的try范围本地，即avg在finally块中不可用。

尝试抓住弄糟的方法

1 个答案: