[R] numerical accuracy, dumb question

[Marc Schwartz]

On Sat, 2004-08-14 at 13:19, Prof Brian Ripley wrote:
> On Sat, 14 Aug 2004, Marc Schwartz wrote:
> 
> > > object.size("a")
> > [1] 44
> > 
> > > object.size(letters)
> > [1] 340
> > 
> > In the second case, as Tony has noted, the size of letters (a character
> > vector) is not 26 * 44.
> 
> Of course not.  Both are character vectors, so have the overhead of any R
> object plus an allocation for pointers to the elements plus an amount for
> each element of the vector (see the end).
> 
> These calculations differ on 32-bit and 64-bit machines.  For a 32-bit
> machine storage is in units of either 28 bytes (Ncells) or 8 bytes
> (Vcells) so single-letter characters are wasteful, viz
> 
> > object.size("aaaaaaa")
> [1] 44
> 
> That is 1 Ncell and 2 Vcells, 1 for the string (7 bytes plus terminator)
> and 1 for the pointer.
> 
> Whereas
> 
> > object.size(letters)
> [1] 340
> 
> has 1 Ncell and 39 Vcells, 26 for the strings and 13 for the pointers 
> (which fit two to a Vcell).
> 
> Note that repeated character strings may share storage, so for example
> 
> > object.size(rep("a", 26))
> [1] 340
> 
> is wrong (140, I think).  And that makes comparisons with factors depend
> on exactly how they were created, for a character vector there probably is 
> a lot of sharing.
> 
> I have a feeling that these calculations are off for character vectors, as 
> each element is a CHARSXP and so may have an Ncell not accounted for by 
> object.size.  (`May' because of potential sharing.)  Would anyone who is 
> sure like to confirm or deny this?
> 
> It ought to be possible to improve the estimates for character vectors a 
> bit as we can detect sharing amongst the elements.

Prof. Ripley,

Thanks for the clarifications. 

I'll need to spend some time reading through R-exts.pdf and
Rinternals.h.

Regards,

Marc