设计模式-简单工厂模式-526互联

用任意一种面向对象语句实现计算机程序，要求输入两个数和符号，得到结果。

小菜：

Console.WriteLine("please input first interger");
string first_number = Console.ReadLine();
Console.WriteLine("please input second interger");
string second_number = Console.ReadLine();
Console.WriteLine("please input yunsaunfu");
string caozuo = Console.ReadLine();

double result_number = 0;

if (caozuo == "+")
result_number = Convert.ToDouble(first_number) + Convert.ToDouble(second_number);

if (caozuo == "-")
result_number = Convert.ToDouble(first_number) + Convert.ToDouble(second_number);

if (caozuo == "*")
result_number = Convert.ToDouble(first_number) + Convert.ToDouble(second_number);

if (caozuo == "/")
result_number = Convert.ToDouble(first_number) + Convert.ToDouble(second_number);

Console.WriteLine(result_number);

问题：

4个if 分支，每个都要走一遍
除法没有考虑异常的情况

改进如下：

Console.WriteLine("please input first interger");
string first_number = Console.ReadLine();

Console.WriteLine("please input second interger");
string second_number = Console.ReadLine();

Console.WriteLine("please input yunsaunfu");
string caozuo = Console.ReadLine();


Console.WriteLine(first_number);
Console.WriteLine(second_number);
Console.WriteLine(caozuo);

double result_number = 0;

switch (caozuo)
{
    case "+":
        result_number = Convert.ToDouble(first_number) + Convert.ToDouble(second_number);
        break;
    case "-":
        result_number = Convert.ToDouble(first_number) - Convert.ToDouble(second_number);
        break;
    case "*":
        result_number = Convert.ToDouble(first_number) * Convert.ToDouble(second_number);
        break;
    case "/":
        if (second_number == "0")
            Console.WriteLine("can't be zero");
        else
            result_number = Convert.ToDouble(first_number) / Convert.ToDouble(second_number);
        break;
}
    
Console.WriteLine(result_number);