Parsing double precision numbers

[marathoner]

After reading the string representation of a double precision
number, -0.417597000000000D+06, from a text file, I am unable to convert it
to a double precision number. Here is the code I am using:

string[] result = block.Split(charSeparators,
StringSplitOptions.RemoveEmptyEntries);
double temp1 = double.Parse(result[0]);

After the last statement, I get an unhandled exception.

Marathoner

 

[Willy Denoyette [MVP]]

After reading the string representation of a double precision
number, -0.417597000000000D+06, from a text file, I am unable to convert
it to a double precision number. Here is the code I am using:

string[] result = block.Split(charSeparators,
StringSplitOptions.RemoveEmptyEntries);
double temp1 = double.Parse(result[0]);

After the last statement, I get an unhandled exception.

Marathoner

-0.417597000000000D+06 is invalid (the D should be an e or E to be valid),
where did you get this from?

Willy.

 

[Tom Porterfield]

After reading the string representation of a double precision
number, -0.417597000000000D+06, from a text file, I am unable to convert it
to a double precision number. Here is the code I am using:

string[] result = block.Split(charSeparators,
StringSplitOptions.RemoveEmptyEntries);
double temp1 = double.Parse(result[0]);

After the last statement, I get an unhandled exception.

What is the D in your number? If that is supposed to mean exponent, the
correct representation is e (or E).

 

[Ignacio Machin \( .NET/ C# MVP \)]

Hi,


The D is incorrect, it should be an E.

Where are you reading these numbers from?

Anyway, use String.Replace
double temp1 = double.Parse(result[0].Replace("D","E"));

 

[Family Tree Mike]

Hi,


The D is incorrect, it should be an E.

Where are you reading these numbers from?

Anyway, use String.Replace
double temp1 = double.Parse(result[0].Replace("D","E"));
--
Ignacio Machin
http://www.laceupsolutions.com
Mobile & warehouse Solutions.
news:[email protected]...

Looks like an old fortran output...