Range of Double

14y2.1kVisual Basic .NET

Range of double is in the book is highlighted in yellow. But program in the following website (http://msdn.microsoft.com/en-us/library/994c0zb1.aspx) says that:

//-1.79769313486232E+308 is outside the range of the Double type.
//1.79769313486232E+308 is outside the range of the Double type.

According to the book double should be:

//-1.79769313486232E+308 is inside the range of the Double type.
//1.79769313486232E+308 is inside the range of the Double type.

Please explain the problem

8 Replies

Know the answer? Post it — somebody with the same question will find it here.

Vulpes14y ago
Accepted answer
The figures in the book aren't quite right.

The range of a double (excluding special values such as Infinity and NaN) is from:

-1.7976931348623157E+308 to 1.7976931348623157E+308

However, due to limitations in the way that numbers are converted to strings, if you apply the ToString() method to Double.MinValue or Double.MaxValue (or even to values 'close' to those extremes) and then use the result in the Double.TryParse method, it will tell you that they're out of range!

You can expect to see this sort of anomaly when dealing with numbers at the extremes of precision which are unimportant for most purposes.
Vulpes14y ago
Well, ideally, you'd like the program to give the correct answers.

In the case of Double.MinValue and Double.MaxValue, this isn't possible because of the way that doubles are converted to strings.

Although I'm not absolutely certain about this, I suspect that the double (in scientific notation) is rounded to a maximum of 15 significant figures and then converted to its string representation.

So, in the case of Double.MaxValue, 1.7976931348623157E+308 is rounded up to 1.79769313486232E+308 which Double.TryParse then says is out of range.

However, when I used 1.79769313486231E+308 (i.e. rounded down) in my previous post, this was then (correctly) reported to be within range.
14y ago
Thank you very much for your help.

In my view code itself is saying double minimum value and maximum value (Double.MinValue.ToString() & Double.MaxValue.ToString()).
Therefore I couldn't see any wrong in saying "inside" as long as we try to find out range of the TYPE double in C# Language. I don't know whether my understanding is correct.
Vulpes14y ago
Sure, try this which should give the correct results commented below:

using System;

namespace asfdas
{
class Program
{
static void Main(string[] args)
{
string value;
double number;

value = "2E+310"; // out of range
if (Double.TryParse(value, out number))
Console.WriteLine(number);
else
Console.WriteLine("{0} is outside the range of a Double.", value);

value = "1.79769313486231E+308"; // within range and should still be when converted
if (Double.TryParse(value, out number))
Console.WriteLine(number);
else
Console.WriteLine("{0} is outside the range of a Double.", value);

Console.ReadKey();
}
}
}

/*
2E+310 is outside the range of a Double.
1.79769313486231E+308
*/
14y ago
Could you please demonstrate with a example program (if possible modify the above program)
Vulpes14y ago
The problem would be that if you presented it with a numeric string that was definitely outside the range of a double (say "2E+310") then it would still tell you that it was within range.
14y ago
So that inside the literal string what is the harm of replacing "outside" by "inside". Following program shows that.

using System;

namespace asfdas
{
class Program
{
static void Main(string[] args)
{
string value;
double number;

value = Double.MinValue.ToString();
if (Double.TryParse(value, out number))
Console.WriteLine(number);
else
Console.WriteLine("{0} is inside the range of a Double.", value);

value = Double.MaxValue.ToString();
if (Double.TryParse(value, out number))
Console.WriteLine(number);
else
Console.WriteLine("{0} is inside the range of a Double.", value);

Console.ReadKey();
}
}
}
/*
-1.79769313486232E+308 is inside the range of a Double.
1.79769313486232E+308 is inside the range of a Double.
*/
14y ago
Thank you very much for your explanation.