On Mon, Jan 5, 2009 at 3:16 PM, Brandon S. Allbery KF8NH
wrote:
> On 2009 Jan 5, at 13:57, David Menendez wrote:
>>
>> 2009/1/5 Ross Mellgren :
>>>
>>> If for some reason you absolutely need to avoid parentheses (mostly as a
>>> thought exercise, I guess), you'd have to have a flipped version of
>
On 2009 Jan 5, at 13:57, David Menendez wrote:
2009/1/5 Ross Mellgren :
If for some reason you absolutely need to avoid parentheses (mostly
as a
thought exercise, I guess), you'd have to have a flipped version of
intercalate:
Or a version of ($) that associates differently.
infixl 0 $$
f $
Thank you everybody!
Vasili
On Mon, Jan 5, 2009 at 12:57 PM, David Menendez wrote:
> 2009/1/5 Ross Mellgren :
> > If for some reason you absolutely need to avoid parentheses (mostly as a
> > thought exercise, I guess), you'd have to have a flipped version of
> > intercalate:
>
> Or a version of
2009/1/5 Ross Mellgren :
> If for some reason you absolutely need to avoid parentheses (mostly as a
> thought exercise, I guess), you'd have to have a flipped version of
> intercalate:
Or a version of ($) that associates differently.
infixl 0 $$
f $$ x = f x
*Main Data.ByteString> :t \x y z ->
In this case you have to use parens -- two dollar signs, like this
B.intercalate $ B.intercalate ByteString [ByteString] $ [ByteString]
would also not type check -- it is exactly equivalent to your first
example:
B.intercalate (B.intercalate ByteString [ByteString] ([ByteString]))
just with
yep ... that is exactly what I meant!! so can I use more $'s or must I use
parens (as you did) to disambiguate?
Vasili
On Mon, Jan 5, 2009 at 12:18 PM, Ross Mellgren wrote:
> Did you mean:
> B.intercalate (B.intercalate ByteString [ByteString]) [ByteString]
>
> ($) applies all the way to the ri
On Mon, Jan 5, 2009 at 10:17 AM, Galchin, Vasili wrote:
> Hi Max,
>
>That is what should happen The inner B.intercalate will produce
> the ByteString to be used by the B.intercalate. ??
>
> Vasili
>
Of course. My mistake. Ross Mellgren seems to be on the money though.
--Max
___
Did you mean:
B.intercalate (B.intercalate ByteString [ByteString]) [ByteString]
($) applies all the way to the right, so you were giving the inner
intercalate two lists of ByteString.
-Ross
On Jan 5, 2009, at 1:17 PM, Galchin, Vasili wrote:
Hi Max,
That is what should happen ...
Hi Max,
That is what should happen The inner B.intercalate will produce
the ByteString to be used by the B.intercalate. ??
Vasili
On Mon, Jan 5, 2009 at 12:13 PM, Max Rabkin wrote:
> 2009/1/5 Galchin, Vasili :
> > Hello,
> >
> > I have the following:
> >
> > B.in
2009/1/5 Galchin, Vasili :
> Hello,
>
> I have the following:
>
> B.intercalate $ B.intercalate
> ByteString
> [ByteString]
> [ByteString]
>
> I get a type error with this. If I comment out the 2nd B.intercalate
> and th
Hello,
I have the following:
B.intercalate $ B.intercalate
ByteString
[ByteString]
[ByteString]
I get a type error with this. If I comment out the 2nd B.intercalate
and the third parameter I get no type errors.
Regar
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