無法訪問 java 中的 while 循環外的變量

Question

在這段代碼中，我試圖找到最多 n 個數字。 我已經聲明了最大外部循環，但以后無法訪問最大變量：

public class MaxOfNnumbers {

    public static void main(String[] args) throws Exception {
        BufferedReader reader = new BufferedReader(new InputStreamReader (System.in));
        int maximum  = Integer.parseInt(reader.readLine ());

        //write your code here
        while(true) {
            String s = reader.readLine ();
            int n = Integer.parseInt ( s );

            if (n > 0) {
                if (n > maximum) {
                    n = maximum;
                }
            }

        }
        System.out.println ( maximum );// Error indicates that , maximum variable is unreachable
    }
}

Answer 1

代碼中的while(true)行設置了一個無限循環。 您需要從該循環中提供退出條件，否則永遠無法達到System.out.println(maximum) 。

Answer 2

這是我為代碼找到的解決方案。

公共 class MaxOfNnumbers {

public static void main(String[] args) throws Exception {
    BufferedReader reader = new BufferedReader(new InputStreamReader (System.in));
    System.out.println ("Enter the amount of numbers to be scanned");


    // if we enter n to be 0 or negative , we cant form an array
    // hence an if condition where if(n > 0) is applied before forming
    // the array .
    int n = Integer.parseInt ( reader.readLine () );
    System.out.println ( "Entered value for n is "+n );

    int leastNum[] = new int[n];
   System.out.println ("Array size is "+leastNum);

    // for reading the data from keyboard least num array is declared
    // this array is fed from the keyboard .
    if(n > 0){
        for(int i = 0; i < n ; i++){
            leastNum[i]=Integer.parseInt ( reader.readLine () );
            System.out.println ("Value stored at "+ i+ "Location is"+leastNum[i]);
        }
         System.out.println ("Least num length is "+leastNum.length);

        int maximum = Integer.MIN_VALUE;

System.out.println("最大值的初始值為"+maximum);

        for(int i =0 ; i < n; i++){

            if(maximum < leastNum[i]){
                maximum = leastNum[i];
            }
        }
        System.out.println(maximum);
    }
}

}