Andrew Bromage:
>Pattern matching on the LHS of a function definitions looks, for all the
>world, like a set of rewrite rules. That's because, in some sense, they
>are.
>
>In this definition:
>
>f (x:xs) = 1 + f xs
>f [] = []
>
>Intuitively, the first rule should only "fire" if the express
G'day all.
Quoting Duncan Coutts <[EMAIL PROTECTED]>:
> Or finally, the "it's what you want most often" argument.
How about the "it's the most natural thing" argument?
Pattern matching on the LHS of a function definitions looks, for all the
world, like a set of rewrite rules. That's because, i
On Tue, 2004-03-30 at 18:01, S. Alexander Jacobson wrote:
> Thanks for the ~ syntax, but my question is really
> why you need it? What benefit do you get from
> "refutable patterns"?
>
> Alternatively, would anything break if a future
> Haskell just treated all patterns as irrefutable?
A short p
A lot. If everything were irrefutable, then the following:
> mymap f (x:xs) = f x : xs
> mymap f [] = []
would never get to the second branch and would fail when it reached the
end of a list. so would it if you put the lines in the other, "more
natural," order, though perhaps it's less clear
Thanks for the ~ syntax, but my question is really
why you need it? What benefit do you get from
"refutable patterns"?
Alternatively, would anything break if a future
Haskell just treated all patterns as irrefutable?
-Alex-
_
S. Al
> "aj" == S Alexander Jacobson <[EMAIL PROTECTED]> writes:
aj> I would assume that this function:
aj> foo list@(h:t) = list
aj> is equivalent to
aj> foo list = list
aj> where (h:t)=list
aj> But passing [] to the first generates an error
aj> even though h
tis 2004-03-30 klockan 17.30 skrev S. Alexander Jacobson:
> I would assume that this function:
>
> foo list@(h:t) = list
>
> is equivalent to
>
> foo list = list
> where (h:t)=list
>
> But passing [] to the first generates an error
> even though h and t are never used! Passing [] to
>
I would assume that this function:
foo list@(h:t) = list
is equivalent to
foo list = list
where (h:t)=list
But passing [] to the first generates an error
even though h and t are never used! Passing [] to
the second works just fine.
At this point, I sort of understand the reason for
M
