我这个函数没走一次就得重新输入一次,怎么把输入放在主函数里然后函数去获取那个值 或者有别的解决办法吗?
public static double tuoqiu()
{
Console.WriteLine("请选择椭球类型:");
Console.WriteLine("如果为克拉索夫斯基椭球请输入1");
Console.WriteLine("如果为IAG-75椭球请输入2");
Console.WriteLine("如果为WGS-84椭球请输入3");
Console.WriteLine("如果为CGCS2000椭球请输入4");
string tuoqiu = Console.ReadLine();
int temp = int.Parse(tuoqiu);
double f=0;
switch (temp)
{
case 1: f = 1 / 298.3; break;
case 2: f = 1 / 298.257; break;
case 3: f = 1 / 298.257223563; break;
case 4: f = 1 / 298.257222101; break;
}
return f;
}
static void Main(string[] args)
{
Console.WriteLine("请选择椭球类型:");
Console.WriteLine("如果为克拉索夫斯基椭球请输入1");
Console.WriteLine("如果为IAG-75椭球请输入2");
Console.WriteLine("如果为WGS-84椭球请输入3");
Console.WriteLine("如果为CGCS2000椭球请输入4");
string tuoqiu = Console.ReadLine();
int temp = int.Parse(tuoqiu);
Console.WriteLine(tuoqiu(temp));
}
public static double tuoqiu(int temp){
double f=0;
switch (temp)
{
case 1: f = 1 / 298.3; break;
case 2: f = 1 / 298.257; break;
case 3: f = 1 / 298.257223563; break;
case 4: f = 1 / 298.257222101; break;
}
return f;
}
你可以这样写:
public static double tuoqiu(int temp)
{
double f = 0;
switch (temp)
{
case 1: f = 1 / 298.3; break;
case 2: f = 1 / 298.257; break;
case 3: f = 1 / 298.257223563; break;
case 4: f = 1 / 298.257222101; break;
}
return f;
}
public static double GetE2(int f)
{
return 2 * f - f * f;
}
使用时:
Console.WriteLine("请选择椭球类型:");
Console.WriteLine("如果为克拉索夫斯基椭球请输入1");
Console.WriteLine("如果为IAG-75椭球请输入2");
Console.WriteLine("如果为WGS-84椭球请输入3");
Console.WriteLine("如果为CGCS2000椭球请输入4");
string str = Console.ReadLine();
int temp = int.Parse(str);
int f = tuoqiu(temp);
int result = GetE2(f);