Question

我正在尝试编写一个Java程序，用户在该程序中输入代表二进制数（0和1）的字符串。程序应验证输入是否为二进制数，如果不是这种情况，则程序将显示一条消息，指出该数字不是有效的二进制数。如果用户输入可接受，则程序应计算该单词中有多少个1，然后输出计数。

这是我到目前为止的内容，但是当我输入“ 1234”之类的字符串并转到第三个while循环时，它只会不断重印消息“您的数字只能包含一和零。请只包含一个二进制值一和零：“，即使我输入的有效字符串如” 0010“。我已经为此花了几天的精力，因此，非常感谢您的协助。

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    String input;


    System.out.println("Welcome to the binary scanner!");
    System.out.print("Please enter a string in binary fashion: ");

    while (!scan.hasNextInt()) {
       System.out.println("That's not a binary number...\nPlease enter a binary number");
       String temp = scan.nextLine();
    }

    input = scan.nextLine();
    input = input.trim();


    while (!input.contains("1")){
        System.out.println("Your number does not contain any ones.\nPlease enter a binary number that contains ones.");
        input = scan.nextLine();            
    }



    while (input.contains("2") || input.contains("3") || input.contains("4") || input.contains("5") || input.contains("6") || input.contains("7") || input.contains("8") || input.contains("9")){
        System.out.println ("Your number can only contain ones and zeros.\nPlease ever a binary value containing only ones and zeros: ");
        input = scan.nextLine();
    }


    while (input.contains("1")){ 
        boolean notOne = false;
        int oneCount = 0;

        for (int x = 0; x < input.length(); x++){
            char c = input.charAt(x);
            if (c == '1' || c == '0'){
                if (c == '1'){
                    oneCount++;
                    notOne = true;
                    }
            }
            else {
                notOne = false;
            }


            }System.out.println("The binary string contains " + oneCount + (oneCount == 1 ? " one." : " ones."));
            break;
    }

}

}

Answer 1

欢迎使用Stackoverflow。我无法重现您的问题（我先输入1234，然后输入0010，效果很好）。但是，由于您的最后一个while循环无用，因此我对您的代码进行了一些修改。请尝试使用“调试器”来识别您的问题，或发布一些屏幕截图以证明您的问题。

public static void main(String[] args)
{
    Scanner scan = new Scanner(System.in);

    System.out.println("Welcome to the binary scanner!");
    System.out.print("Please enter a string in binary fashion: ");

    while (!scan.hasNextInt())
    {
        System.out.println("That's not a binary number...\nPlease enter a binary number");
        String temp = scan.nextLine();
    }

    String input = scan.nextLine();
    input = input.trim();


    while (!input.contains("1"))
    {
        System.out.println("Your number does not contain any ones.\nPlease enter a binary number that contains ones.");
        input = scan.nextLine();
    }


    while (input.contains("2") || input.contains("3") || input.contains("4") || input.contains("5") || input.contains("6") || input.contains("7") || input.contains("8") || input.contains("9"))
    {
        System.out.println("Your number can only contain ones and zeros.\nPlease ever a binary value containing only ones and zeros: ");
        input = scan.nextLine();
    }


    int oneCount = 0;

    for (int x = 0; x < input.length(); x++)
    {
        char c = input.charAt(x);
        if (c == '1')
        {
            oneCount++;
        }
    }
    System.out.println("The binary string contains " + oneCount + (oneCount == 1 ? " one." : " ones."));

}

Java-无法突破while循环

1 个答案: