尝试double.Parse（）时发生System.Format.FormatException

Question

My program throws a System.FormatException: "The entrystring has the wrong format." whenever I am trying run this code:

public double[] ReturnCoordsFromString(string CoordString)
    {
        string[] cArrStr = CoordString.Split(' ');

        List<double> NumList = new List<double>();

        foreach (string elem in cArrStr)
        {
            Console.WriteLine(elem);
            double b = Convert.ToDouble(elem);   // <= Error is here
            NumList.Add(b);
        }
        double[] retN = NumList.ToArray<double>();

        return retN;
    }

I have also tried to run it with Convert.ToDouble(elem) and Encodings with ascii and utf_8. None of these worked.

To understand my code:

I call the function from another function and the CoordString argument looks like this: 90 10 1000 So they are all Integers, but I need them as double. (I tried Int32.Parse() and then convert to double, here it crashes on the Int32.Parse() part)

My code should get the CoordString ( "90 10 1000" ) and split it into single strings ( ["90", "10", "1000"] ). The Console.WriteLine(elem) prints the correct numbers, no letters, just numbers as string.

Any idea why / how to fix it? Nothing other questions suggested worked so far.

EDIT:

The weird thing is, printing elem does work well. But the Exception Window shows me this:

b        0         double
elem     ""        string
// The class name here

Answer 1

You probably have a double space somewhere that's causing the problem. Try specifying StringSplitOptions.RemoveEmptyEntries :

string[] cArrStr = CoordString(' ', StringSplitOptions.RemoveEmptyEntries)

instead of

string[] cArrStr = CoordString.Split(' ');

Also, you should use Double.TryParse and not Convert.ToDouble , since Double.TryParse will only return false when it can't convert, while Convert.ToDouble will throw an exception:

So use this:

double b;
if(Double.TryParse(elem, out d))
{
    // value is a double
}

instead of

double b = Convert.ToDouble(elem);