After reading the integer and the newline character (which is just ignored), the readPosInt method tests the integer. If the integer is bigger than or equal to zero, then the readPosInt method returns the integer as result. If the integer is strictly less than zero, then the readPosInt method uses System.out.println to print the error message "Positive integers only!" to the user (use System.out.println for this, not System.err.println, otherwise you might hit a bug in Eclipse...), and then does the whole thing again (including printing again the string argument of the readPosInt method) to try to read an integer again (hint: just print the error message, and then the while loop you already have around the whole code will automatically do the whole thing again...)接上面的代码

这段代码可能是一个方法的实现，名为readPosInt。根据你的描述，这个方法读取用户输入的整数，如果这个整数为非负数，则直接返回这个整数；否则，向用户输出“Positive integers only!”的错误提示信息，然后重新读取整数。这个过程会一直重复，直到读取到一个非负整数为止。

以下是可能的readPosInt方法的实现：

public static int readPosInt(String prompt) {
    Scanner scanner = new Scanner(System.in);
    int num;
    do {
        System.out.print(prompt);
        num = scanner.nextInt();
        if (num < 0) {
            System.out.println("Positive integers only!");
        }
    } while (num < 0);
    return num;
}

注：这个实现假设用户输入的是合法的整数，如果用户输入非整数或超出int类型的取值范围，会抛出异常。为了让程序更健壮，可以加上对这些异常的处理。

相关问题

After reading the integer, the readPosInt method must also use the scanner’s nextLine method to read the single newline character that comes from the user pressing the Enter key on the keyboard after typing the integer (if you do not read this newline character using the nextLine method inside the readPosInt method, then the newline character will remain in the input stream, and, the next time you use the readLine method described above, the readLine method will just immediately read only the newline character from the input stream and return an empty string as result, without waiting for the user to type anything!写一个java文件

Sure, here is an example Java file that implements the readPosInt method with the described behavior:

import java.util.Scanner;

public class ReadPosIntExample {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int num = readPosInt(scanner);
        System.out.println("You entered: " + num);
        scanner.close();
    }

    public static int readPosInt(Scanner scanner) {
        int num = 0;
        boolean validInput = false;
        while (!validInput) {
            System.out.print("Enter a positive integer: ");
            if (scanner.hasNextInt()) {
                num = scanner.nextInt();
                scanner.nextLine(); // consume newline character
                if (num > 0) {
                    validInput = true;
                } else {
                    System.out.println("Error: Input must be positive.");
                }
            } else {
                System.out.println("Error: Input must be an integer.");
                scanner.nextLine(); // consume invalid input line
            }
        }
        return num;
    }
}

This program prompts the user to enter a positive integer, reads an integer value using the scanner's nextInt method, and then uses the scanner's nextLine method to consume the newline character that follows the integer input. If the input is not a positive integer, the program displays an error message and asks the user to input again. The program continues to prompt the user until a valid positive integer is entered. Finally, the program prints the entered integer to the console.

Print the integers from 1 to 20 using a while loop and the int counter variable x. Print only 5 integers per line. [Hint: When x % 5 is 0, print a newline character; otherwise, print a tab character.]

Here's the code to achieve the task:

int x = 1;
while (x <= 20) {
  cout << x << "\t";
  if (x % 5 == 0) {
    cout << endl; // print a newline character when x is a multiple of 5
  }
  x++;
}

Output:

1	2	3	4	5	
6	7	8	9	10	
11	12	13	14	15	
16	17	18	19	20