Question

第一次发布，并对是否带有方法的语句有疑问。

我一直在尝试多种方法来做到这一点，但是还没有使它起作用。我正在尝试根据用户是否键入1到6之间的数字来调用方法。这是我到目前为止的结果，请不要判断计算器的对话框（我只有14岁）。

 class Program
{
    static void Main(string[] args)
    {

        Console.WriteLine(Work());
    }

    public static void Work(int input)
    {
        if (input < 0)
        {
            Console.WriteLine("Please enter a number between 1 and 6");
        }
        else if (input == 1)
        {
            Console.WriteLine(Add());
        }
        else if (input == 2)
        {
            Console.WriteLine(Subtract());
        }



    }

    public static int Add()
    {
        Console.WriteLine("Hey bro, need a new number man!");
        string input1 = Console.ReadLine();
        Console.WriteLine("Gnarly dude, how 'bout a second one?");
        string input2 = Console.ReadLine();
        Console.WriteLine("Here botine-shake, there is your final number.");

        int num1 = int.Parse(input1);
        int num2 = int.Parse(input2);

        int result = num1 + num2;
        return result;
    }

    public static int Subtract()
    {
        Console.WriteLine("Number. Now. Please hurry.");
        string input1 = Console.ReadLine();
        Console.WriteLine("Need another number. Hurry.");
        string input2 = Console.ReadLine();
        Console.WriteLine("Here is your number. Now please leave.");

        int num1 = int.Parse(input1);
        int num2 = int.Parse(input2);

        int result = num1 - num2;
        return result;
    }

    public static int Multiply()
    {
        Console.WriteLine("Gimme that number.");
        string input1 = Console.ReadLine();
        Console.WriteLine("Ok, how would you like to give me another?");
        string input2 = Console.ReadLine();
        Console.WriteLine("Here you go baby.");

        int num1 = int.Parse(input1);
        int num2 = int.Parse(input2);

        int result = num1 * num2;
        return result;
    }

    public static decimal Divide()
    {
        Console.WriteLine("/enter.Num1");
        string input1 = Console.ReadLine();
        Console.WriteLine("/enter.Num2");
        string input2 = Console.ReadLine();
        Console.WriteLine("/final.Dividend");

        decimal num1 = decimal.Parse(input1);
        decimal num2 = decimal.Parse(input2);

        decimal result = num1 / num2;
        return result;
    }

    public static int Square()
    {
        Console.WriteLine("What number do you want SQUARED?");
        string input1 = Console.ReadLine();
        Console.WriteLine("Here's your square!");


        int num1 = int.Parse(input1);
        int result = num1 * num1;
            return result;
    }

    public static int Cube()
    {
        Console.WriteLine("What have thine want CUBED?");
        string input1 = Console.ReadLine();
        Console.WriteLine("Here, I bestow upon you your cube...");

        int num1 = int.Parse(input1);
        int result = num1 * num1 * num1;
        return result;
    }





}

Answer 1

我不太明白您的意思，但是应该修复一些问题：

static void Main(string[] args)
{
    Console.WriteLine(Work());
}
public static void Work(int input)
......

您可以看到Work()是一个“动作”，而不是一个“功能”。由于它不返回任何内容，因此不适合WriteLine的结果Work()

顺便说一句，您可以看到Work(int input)需要输入，显然您没有提供。试试这个：

static void Main(string[] args)
{
    int myInput = int.Parse(Console.ReadLine());
    Work(myInput);
}
public static void Work(int input)
{ .....

Answer 2

根据您提供的代码，看起来您在调用Work方法之前没有要求用户输入，因此不会发生任何事情，因为没有参数传递给Work方法。

您应该尝试这样的事情：

static void Main(string[] args)
    CaptureUserInput();
}

public static void WriteInstructions() {
    Console.WriteLine("Enter a number between 1 and 6 to perform a calculation");
    // instruct the user for each available operation
    Console.WriteLine("1: Addition");
    Console.WriteLine("2: Subtraction");
}

public static void CaptureUserInput() {
    WriteInstructions();

    // capture the user's input and convert it to an integer
    string stringInput = Console.Readline();
    int input = int.Parse(stringInput);

    // validate that it is a valid integer
    if (Enumerable.Range(1,6).Contains(input)) {
        // this is a valid number in the range we want, call the Work method
        Work(input);
    } else {
        // the user has entered an invalid entry, prompt them and wait for another attempt
        Console.WriteLine("Sorry, that is an invalid option.");
        CaptureUserInput();
    }
}

请注意，此设置是为了使您可以在用户输入的数字不是数字或超出1-6的范围时递归处理。这将提示用户输入的内容无效，然后等待下一次尝试输入一个数字。

如果语句和方法

2 个答案: