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From: Maik Beckmann (maikbeckmann_at_[hidden])
Date: 2008-01-26 06:32:55


Am Samstag 26 Januar 2008 12:06:34 schrieb Sebastian Gesemann:

> 127^5 * 5 doubles need
> 2^(log2(127)*5 + log2(5 columns) + log2(8 bytes/double)) bytes = 1.2 Tera
> Bytes

Or simply:
  ( (127^5)*5)*8 / (1024^3)
  =
    1230.78 GByte