[Rd] Confused about NAMED

[Matthew Dowle]

>
> On Nov 24, 2011, at 11:13 , Matthew Dowle wrote:
>
>> Hi,
>>
>> I expected NAMED to be 1 in all these three cases. It is for one of
>> them,
>> but not the other two?
>>
>>> R --vanilla
>> R version 2.14.0 (2011-10-31)
>> Platform: i386-pc-mingw32/i386 (32-bit)
>>
>>> x = 1L
>>> .Internal(inspect(x))   # why NAM(2)? expected NAM(1)
>> @2514aa0 13 INTSXP g0c1 [NAM(2)] (len=1, tl=0) 1
>>
>>> y = 1:10
>>> .Internal(inspect(y))   # NAM(1) as expected but why different to x?
>> @272f788 13 INTSXP g0c4 [NAM(1)] (len=10, tl=0) 1,2,3,4,5,...
>>
>>> z = data.frame()
>>> .Internal(inspect(z))   # why NAM(2)? expected NAM(1)
>> @24fc28c 19 VECSXP g0c0 [OBJ,NAM(2),ATT] (len=0, tl=0)
>> ATTRIB:
>>  @24fc270 02 LISTSXP g0c0 []
>>    TAG: @3f2120 01 SYMSXP g0c0 [MARK,gp=0x4000] "names"
>>    @24fc334 16 STRSXP g0c0 [] (len=0, tl=0)
>>    TAG: @3f2040 01 SYMSXP g0c0 [MARK,gp=0x4000] "row.names"
>>    @24fc318 13 INTSXP g0c0 [] (len=0, tl=0)
>>    TAG: @3f2388 01 SYMSXP g0c0 [MARK,gp=0x4000] "class"
>>    @25be500 16 STRSXP g0c1 [] (len=1, tl=0)
>>      @1d38af0 09 CHARSXP g0c2 [MARK,gp=0x21,ATT] "data.frame"
>>
>> It's a little difficult to search for the word "named" but I tried and
>> found this in R-ints :
>>
>>    "Note that optimizing NAMED = 1 is only effective within a primitive
>> (as the closure wrapper of a .Internal will set NAMED = 2 when the
>> promise to the argument is evaluated)"
>>
>> So might it be that just looking at NAMED using .Internal(inspect()) is
>> setting NAMED=2?  But if so, why does y have NAMED==1?
>
> This is tricky business... I'm not quite sure I'll get it right, but let's
> try
>
> When you are assigning a constant, the value you assign is already part of
> the assignment expression, so if you want to modify it, you must
> duplicate. So NAMED==2 on z <- 1 is basically to prevent you from
> accidentally "changing the value of 1". If it weren't, then you could get
> bitten by code like for(i in 1:2) {z <- 1; if(i==1) z[1] <- 2}.
>
> If you're assigning the result of a computation, then the object only
> exists once, so
> z <- 0+1  gets NAMED==1.
>
> However, if the computation is done by returning a named value from within
> a function, as in
>
>> f <- function(){v <- 1+0; v}
>> z <- f()
>
> then again NAMED==2. This is because the side effects of the function
> _might_ result in something having a hold on the function environment,
> e.g. if we had
>
> e <- NULL
> f <- function(){e <<-environment(); v <- 1+0; v}
> z <- f()
>
> then z[1] <- 5 would change e$v too. As it happens, there aren't any side
> effects in the forme case, but R loses track and assumes the worst.
>

Thanks a lot, think I follow. That explains x vs y, but why is z NAMED==2?
The result of data.frame() is an object that exists once (similar to 1:10)
so shouldn't it be NAMED==1 too?  Or, R loses track and assumes the worst
even on its own functions such as data.frame()?

>>
>> Thanks!
>> Matthew
>>
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>
> --
> Peter Dalgaard, Professor
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
> Phone: (+45)38153501
> Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com
>
>