使用字符串的 C# 计算器 [关闭]

Question

我需要在不使用任何库的情况下解决字符串行上的算术运算。我已经尝试了以下代码-

string s = Console.ReadLine();

switch (s[1])
{
    case s[1] == "+":
        Console.WriteLine("Result: " + (s[0] + s[2]));
        break;
    case "-":
        Console.WriteLine("Result: " + (s[0] - s[2]));
        break;
    case "*":
        Console.WriteLine("Result: " + (s[0] * s[2]));
        break;
    case "/":
        if (s[2] != 0)
            Console.WriteLine("Result: " + (s[0] / s[2]));
        else
            Console.WriteLine("Indeterminate");
        break;
}

Console.WriteLine(s);

但我不确定如何将字符串转换为字符。

有什么办法可以使这个工作吗？我正在寻找任何建议或提示。

Answer 1

编译器告诉你问题：

 CS0029 Cannot implicitly convert type 'string' to 'char'   

此错误发生在case 语句中。它需要char，而不是string。 C# 中的字符串文字用双引号括起来，字符文字（char）用单引号括起来。

在第一个 case 语句中，您有 s[1] == "+"。那里不应该有比较（switch case 会为你做这件事）

switch (s[1]) // s[1] is of type char 
{
    case '+': 
        Console.WriteLine("Result: " + (s[0] + s[2]));
        break;
    case '-':
        Console.WriteLine("Result: " + (s[0] - s[2]));
        break;
    case '*':
        Console.WriteLine("Result: " + (s[0] * s[2]));
        break;
    case '/':
        if (s[2] != 0)
            Console.WriteLine("Result: " + (s[0] / s[2]));
        else
            Console.WriteLine("Indeterminate");
        break;
}

最后，您应该注意字符文字数字的数值与其表示不同。您应该解析字符以获取数值。例如：

var value1 = int.Parse(s[0].ToString());
var value2 = int.Parse(s[2].ToString());

接下来，您将意识到您的计算器仅支持 1 位数字，您可能需要预先添加一些输入验证以避免程序因无效输入而崩溃。

Answer 2

正如 jeroenh 提到的，问题在于您在 switch 情况下将 char 与 string 进行比较，这是不允许的。您必须比较相似的类型。

而且，最好先将输入字符串解析为正确的类型，然后进行比较和数学运算，而不是依赖字符串中字符的索引。这使您可以更灵活地进行用户输入，以防他们输入多位数字，或在运算符周围添加空格。

例如：

// These are the allowed operators
var operators = new[] {'+', '-', '/', '*'};

// Get user input
string s = Console.ReadLine();

// Find the first index of an operator
var operatorIndex = operators.Any(s.Contains)
    ? operators.Select(op => s.IndexOf(op)).Where(index => index > -1).Min()
    : -1;

// If no valid operators are present, give an error message to the user
if (operatorIndex < 0)
{
    throw new ArgumentException(
        $"Input doesn't contain a valid operator ({string.Join(", ", operators)}).");
}

// Based on the operator position, find the portions of the string that
// represent the first number, the operator, and the second number.
var firstNumber = s.Substring(0, operatorIndex);
var operatorChar = s[operatorIndex];
var secondNumber = s.Substring(operatorIndex + 1);
        
int operand1, operand2;

// Try to parse the first and second number to integers. If either 
// of them are invalid, give an error message to the user
if (!int.TryParse(firstNumber, out operand1) ||
    !int.TryParse(secondNumber, out operand2))
{
    throw new ArgumentException(
        $"Input doesn't contain a valid integer on each side of the operator.");
}

// Now we can swich on the operator character and do math with the integers
switch (operatorChar)
{
    case '+':
        Console.WriteLine("Result: " + (operand1 + operand2));
        break;
    case '-':
        Console.WriteLine("Result: " + (operand1 - operand2));
        break;
    case '*':
        Console.WriteLine("Result: " + (operand1 * operand2));
        break;
    case '/':
        if (operand2 != 0)
            Console.WriteLine("Result: " + (operand1 / operand2));
        else
            Console.WriteLine("Indeterminate");
        break;
}

Console.Write("\nPress any key to exit...");
Console.ReadKey();

Answer 3

如果任务不排除 .Net 中可用的命名空间，您可以根据您的需要调整 Microsoft DotNet Documentation 中的这个奇妙示例。

这是他们解释的代码的 sn-p：

String[] expressions = {
 "16 + 21",
 "31 * 3", 
 "28 / 3",
 "42 - 18", 
 "12 * 7",
 "2, 4, 6, 8" };

String pattern = @"(\d+)\s+([-+*/])\s+(\d+)";

foreach (var expression in expressions)
  foreach (System.Text.RegularExpressions.Match m in System.Text.RegularExpressions.Regex.Matches(expression, pattern)) 
  {
    int value1 = Int32.Parse(m.Groups[1].Value);
    int value2 = Int32.Parse(m.Groups[3].Value);
    switch (m.Groups[2].Value)
    {
      case "+":
        Console.WriteLine("{0} = {1}", m.Value, value1 + value2);
        break;
      case "-":
        Console.WriteLine("{0} = {1}", m.Value, value1 - value2);
        break;
      case "*":
        Console.WriteLine("{0} = {1}", m.Value, value1 * value2);
        break;
      case "/":
        Console.WriteLine("{0} = {1:N2}", m.Value, value1 / value2);
        break;
    }
  }
// The example displays the following output:
//       16 + 21 = 37
//       31 * 3 = 93
//       28 / 3 = 9.33
//       42 - 18 = 24
//       12 * 7 = 84

使用正则表达式，您可以“保存”原始输入字符串的各个部分，并将其分解为表达式的匹配值。

这是正则表达式的作用的图像：

正则表达式将您的字符串分解为您的第一个数字的捕获组（可通过m.Groups[1] 获得并保存到value1），您的操作员的捕获组到组 2 和您的第二个数字的捕获组到第 3 组。

采用Console.ReadLine() 输入并使用正则表达式可以得到类似的结果：

using System.Text.RegularExpressions;


//...

string expression = Console.ReadLine()

String pattern = @"(\d+)\s+([-+*/])\s+(\d+)";
 
foreach (Match m in Regex.Matches(expression, pattern)) 
{
  // As others have pointed out you need to parse your numbers:
  int value1 = Int32.Parse(m.Groups[1].Value);
  int value2 = Int32.Parse(m.Groups[3].Value);
  
  // Now we switch case the operator in Group 2
  switch (m.Groups[2].Value)
  {
    case "+":
      Console.WriteLine("{0} = {1}", m.Value, value1 + value2);
      break;
    case "-":
      Console.WriteLine("{0} = {1}", m.Value, value1 - value2);
      break;
    case "*":
      Console.WriteLine("{0} = {1}", m.Value, value1 * value2);
      break;
    case "/":
      Console.WriteLine("{0} = {1:N2}", m.Value, value1 / value2);
      break;
  }
}