On 4/1/2009 10:38 AM, Stavros Macrakis wrote:
NOTA BENE: This email is about `=`, the assignment operator (e.g. {a=1}
which is equivalent to { `=`(a,1) } ), not `=` the named-argument syntax
(e.g. f(a=1), which is equivalent to
eval(structure(quote(f(1)),names=c('','a'))).
As far as I can tell from the documentation, assignment with = is precisely
equivalent to assignment with <-. Yet they call different primitives:
The parser does treat them differently:
> if (x <- 2) cat("assigned\n")
assigned
> if (x = 2) cat("assigned\n")
Error: unexpected '=' in "if (x ="
The ?"=" man page explains this:
" The operator '<-' can be used anywhere,
whereas the operator '=' is only allowed at the top level (e.g.,
in the complete expression typed at the command prompt) or as one
of the subexpressions in a braced list of expressions. "
though the restriction on '=' seems to be described incorrectly:
> if ((x = 2)) cat("assigned\n")
assigned
in which the assignment is in parentheses, not a braced list.
As to the difference between the operations of the two primitives: see
do_set in src/main/eval.c. The facility is there to distinguish between
them, but it is not used.
Duncan Murdoch
`=`
.Primitive("=")
`<-`
.Primitive("<-")
(Perhaps these are different names for the same internal function?)
Also, the difference is preserved by the parser:
quote({a=b})
{
a = b
}
quote({a<-b})
{
a <- b
}
even though in other cases the parser canonicalizes variant syntax, e.g. ->
to <-:
quote({a->b})
{
b <- a
}
`->`
Error: object "->" not found
Is there in fact some semantic difference between = and <- ?
If not, why do they use a different operator internally, each calling a
different primitive?
Or is this all just accidental inconsistency resulting from the '=', '<-',
and '->' features being added at different times by different people working
off different stylistic conventions?
-s
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