Get a Round-tripped Double with only 3 decimals and +/- symbol

Question

I need to convert a Double into a String respecting some conditions:

• I don't want a rounded value
• Only 3 digits after comma
• + or - symbol and beginning.
• (Extra : 0.000 if the double value is 0) (Not the most important)

So, if my Double value is 0.257894214 I want the string to be : +0.257.

I tested a few lines of code :

Double num = 0.257894214;
Double num2 = -0.257894214;
Double num3 = 0;

num.ToString("R")   // 0,257894214
num2.ToString("R")  // -0,257894214
num3.ToString("R")  // 0

String.Format("{0:0.000}",num)      // 0,258
String.Format("{0:0.000}",num2)     // -0,258
String.Format("{0:0.000}",num3)     // 0,000
num.ToString("0.000")               // 0,258
num2.ToString("0.000")              // -0,258
num3.ToString("0.000")              // 0,000

num.ToString("+0.000;-0.000;0.000")             // +0,258
num2.ToString("+0.000;-0.000;0.000")            // -0,258
num3.ToString("+0.000;-0.000;0.000")            // 0,000
String.Format("{0:+0.000;-0.000;0.000}",num)    // +0,258
String.Format("{0:+0.000;-0.000;0.000}",num2)   // -0,258
String.Format("{0:+0.000;-0.000;0.000}",num3)   // 0,000

I feel near the solution but i'm now completely lost!

The last lines are really near what i'm searching for ... but the Double value is rounded ... :(

I hope I have been clear. :) Thanks in advance.

Au revoir !

Answer 1

If I've understood you're post, what you're saying is that you want to truncate at the 3rd decimal place rather than round. So you get the values +0.257, -0.257 and 0.000

The following should do that:

First define a function

// truncates to 3 decimal places
static double Truncate(double num)
{
    return Math.Truncate(num * 1000) / 1000.0;
}

Now try

x = String.Format("{0:+0.000;-0.000;0.000}", Truncate(num));
y = String.Format("{0:+0.000;-0.000;0.000}", Truncate(num2));
z = String.Format("{0:+0.000;-0.000;0.000}", Truncate(num3));

This should work for both positive and negative numbers.