请求实现：输入1-99数字，2次失败后执行system.close(0)

Got it, let's break this down into a practical implementation. The core requirements are: validate user input is between 1-99, track consecutive failed attempts, and exit the program after 2 failures. I'll use Python first since it's readable and widely used, then add a Java example for reference.

Python Implementation #

import sys

def get_valid_number():
    failed_attempts = 0
    while True:
        user_input = input("Please enter a number between 1 and 99: ")
        try:
            number = int(user_input)
            if 1 <= number <= 99:
                print(f"Valid number entered: {number}")
                return number
            else:
                print("Error: Number must be between 1 and 99.")
                failed_attempts += 1
        except ValueError:
            print("Error: Please enter a valid integer.")
            failed_attempts += 1
        
        # Check if we've hit the failure limit
        if failed_attempts >= 2:
            print("Too many failed attempts. Exiting program.")
            sys.exit(0)

# Call the function to test
get_valid_number()

Key Details: #

• We initialize failed_attempts to 0 to track consecutive errors
• The try-except block handles cases where the user enters non-integer values (like text)
• Each invalid input (out of range or non-numeric) increments the failure counter
• Once failed_attempts reaches 2, we print a message and call sys.exit(0) to close the program

Java Implementation #

If you're working with Java, here's an equivalent version:

import java.util.Scanner;

public class InputValidator {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int failedAttempts = 0;
        
        while (true) {
            System.out.print("Please enter a number between 1 and 99: ");
            String userInput = scanner.nextLine();
            
            try {
                int number = Integer.parseInt(userInput);
                if (number >= 1 && number <= 99) {
                    System.out.println("Valid number entered: " + number);
                    scanner.close();
                    return;
                } else {
                    System.out.println("Error: Number must be between 1 and 99.");
                    failedAttempts++;
                }
            } catch (NumberFormatException e) {
                System.out.println("Error: Please enter a valid integer.");
                failedAttempts++;
            }
            
            if (failedAttempts >= 2) {
                System.out.println("Too many failed attempts. Exiting program.");
                scanner.close();
                System.exit(0);
            }
        }
    }
}

Notes for Java: #

• We use Scanner to read user input
• NumberFormatException catches non-integer inputs
• After 2 failures, System.exit(0) terminates the program as requested

Let me know if you need this adapted to another language—this logic works across most programming languages with basic input handling and conditional checks.

内容的提问来源于stack exchange，提问作者Big B