Cannot implicitly convert type 'double' to

问题:

In this code i got the above error in lines i commented.

public double bigzarb(long u, long v)
{
    double n;
    long x;
    long y;
    long w;
    long z;
    string[] i = textBox7.Text.Split(',');
    long[] nums = new long[i.Length];
    for (int counter = 0; counter < i.Length; counter++)
    {
        nums[counter] = Convert.ToInt32(i[counter]);
    }

    u = nums[0];
    int firstdigits = Convert.ToInt32(Math.Floor(Math.Log10(u) + 1));
    v = nums[1];
    int seconddigits = Convert.ToInt32(Math.Floor(Math.Log10(v) + 1));
    if (firstdigits >= seconddigits)
    {
        n = firstdigits;

    }
    else
    {
        n = seconddigits;        
    }
    if (u == 0 || v == 0)
    {
        MessageBox.Show("the Multiply is 0");
    }

    int intn = Convert.ToInt32(n);
    if (intn <= 3)
    {
        long uv = u * v;
        string struv = uv.ToString();
        MessageBox.Show(struv);
        return uv;
    }
    else
    {
        int m =Convert.ToInt32(Math.Floor(n / 2));

        x = u % Math.Pow(10, m); // here
        y = u / Math.Pow(10, m); // here
        w = v % Math.Pow(10, m); // here
        z = v / Math.Pow(10, m); // here

        long result = bigzarb(x, w) * Math.Pow(10, m) + (bigzarb(x, w) + bigzarb(w, y)) * Math.Pow(10, m) + bigzarb(y, z);///here
        textBox1.Text = result.ToString();
        return result;
    }
}

Whats is the problem? Thanks!

回答1:

The Math.Pow method returns a double, not a long so you will need to change your code to account for this:

x = (long)(u % Math.Pow(10, m));

This code will cast the double result from Math.Pow and assign that value to x. Keep in mind that you will lose all the precision providided by decimal (which is a floating-point type and can represent decimal values). Casting to long will truncate everything after the decimal point.

回答2:

Math.Pow returns a double.

the Right Hand Side (RHS) of % can only be an integer type.

you need

x = u % (long)Math.Pow(10, m);///<----here
y = u / (long)Math.Pow(10, m);///here
w = v % (long)Math.Pow(10, m);///here
z = v / (long)Math.Pow(10, m);///here

Additionally, You have the possibility of dividing by zero and destroying the universe.

回答3:

Math.Pow returns a double. You could explicitly cast to long, for example

x = u % (long)Math.Pow(10, m);

although that is likely not the correct solution. Are you certain that the results that you are after can be properly expressed as a double? If not then change the variables to be declared as doubles rather than longs.

回答4:

Change types

long x;
long y;
long w;
long z; 

to

double x;
double y;
double w;
double z; 

Or make use of

Convert.ToInt64

回答5:

You cant' cast implicitly double to long, use (long) cast or change type of variable declaration to double.

回答6:

Also you can use this:

Convert.ToInt64( u % Math.Pow(10, m) )

Source here