This solves my problem. Thank you very much!
Adam On Thu, 2005-12-15 at 12:21 +0000, Paulo Soares wrote: > ByteBuffer.formatDouble() is designed to be fast and output 5 > significative digits, not 5 decimal digits in all cases. This compromise > works well but I also realize that for the odd cases more precision may > be needed. There's version in the CVS that will always output 6 decimal > digits if ByteBuffer.HIGH_PRECISION = true. > > Paulo >
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