The indentation on the second line would generate a parse error, the same as it does now. What parser error is that? Both
function 0 = 0 where fun 1 = 1 function 2 = 2 and function 0 = 0 where fun 1 = 1 fun 2 = 2 works for me. /J On 23 August 2010 11:46, John Smith <voldermort@hotmail.com> wrote:
The indentation on the second line would generate a parse error, the same as it does now.
On 23/08/2010 12:32, Jonas Almström Duregård wrote:
Maybe because of this:
function 0 = 0 where fun 1 = 1 2 = 2
The last declaration (2=2) can define either fun or function. I'm not saying this is a major problem, but there may be other problems like these.
/J
On 23 August 2010 11:15, Brent Yorgey <byorgey@seas.upenn.edu <mailto: byorgey@seas.upenn.edu>> wrote:
On Mon, Aug 23, 2010 at 09:33:13AM +0300, John Smith wrote:
Why doesn't Haskell allow something like this?
fac 0 = 0 1 = 1 x = x * fac (x-1)
This would be clearer than repeating the function name each time, and follow the same pattern as guards and case.
Good question. I don't know of any particular reason.
-Brent
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