Wacek Kusnierczyk wrote:
>
> a simplified example may help to get a clear picture:
>
> i = 1; y = 1:3;
> (while(TRUE) {
> y[i] = 0
> if (i == 2) break
> i = i + 1
> y })
> # 0 0 3
>
> i = 1; y = 1:3;
> (while(TRUE) {
> y[i] = 0
> if (i == 2) break
> i = i + 1
> y + 0 })
> # 0 2 3
>
> the test on i is done after the assignment to y[i]. when the loop
> breaks, y is 0 0 3, and one might expect this to be the final result.
> it looks like the result is the value of y from the previous iteration,
> and it does not seem particularly intuitive to me. (using common sense,
> i mean; an informed expert on the copy-when-scared semantics may have a
> different opinion, but why should a casual user ever suspect such magic.)
>
> anyway, i'd rather expect NULL to be returned. for the oracle,
> ?'while', says:
>
> "'for', 'while' and 'repeat' return the value of the last expression
> evaluated (or 'NULL' if none was), invisibly. [...] 'if' returns the
> value of the expression evaluated, or 'NULL' if none was. [...] 'break'
> and 'next' have value 'NULL', although it would be strange to look for a
> return value."
>
> when i is 2, i == 2 is TRUE. hence, if (i == 2) break evaluates to
> break. break evaluates to NULL, breaks the loop, and the return value
> should be NULL. while it is, following the docs, strange to have q =
> while(...) ... in the code, the result above is not compliant with the
> docs at all -- seems like a plain bug. there is no reason for while to
> return the value of y, be it 0 0 3 or 0 2 3.
>
somewhat surprising to learn,
i = 1
y = 1:3
(while (TRUE) {
y[i] = 0
if (i == 2) { 2*y; break }
i = i + 1
y })
# 0 0 3
where clearly the last expression evaluated (before the break, that is)
is 2*y -- or?
vQ
______________________________________________
R-devel@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
