Because both bit0 and bit1 are free *local* variables
within the case expression. So, they have nothing
to do with your defined functions bit0 and bit1.
Best regards,
Salvador.
[EMAIL PROTECTED] wrote:
Can a kind soul please enlighten me on why f bit0 and f bit1
both return 0?
bit0 =
Dear Martin,
I think that the (practical) reason is avoiding equality checks during
pattern matching.
For instance, how do you evaluate this:
foo ((+1):(1+):[])?
Both expressions in the first and second entries of the list are
semantically equivalent,
but from an operational point of
Dear Peter,
I recently had a similar problem with the same Haskell Platform version
but on a Macmini (running Lion as well).
I learnt that you have to install the 'command line tools' by using the
'preferences' of your XCode installation. After that, everything will work.
Best regards,
Hi,
There is a workshop on Functional Programming and also
a conference on programming languages
http://babel.ls.fi.upm.es/tpf2013/cfp_english.txt
If interested, you can contact the organizers...
Regards,
Salvador.
El 06/09/13 22:59, Joachim Breitner escribió:
Hi,
I'll be visiting
