On 18/07/2013 18:36, Mark Thomas wrote:
> On 18/07/2013 18:24, Violeta Georgieva wrote:
>> 2013/7/18 Mark Thomas wrote:
>>> I look forward to your next brain teaser :)
>>>
>>
>> ;)
>>
>> If I modify the example above like this
>>
>> f = ()->y->2-y; f()(1)
>
> Thanks for all the testing.
>
> I'll
On 18/07/2013 18:24, Violeta Georgieva wrote:
> 2013/7/18 Mark Thomas wrote:
>>
>> On 18/07/2013 08:00, Violeta Georgieva wrote:
>>> Hi,
>>>
>>> Let's have a lambda expression:
>>>
>>> x->y->x-y
>>>
>>> Let's now assign it and invoke it indirectly:
>>>
>>> f = x->y->x-y; f(2)(1)
>>
>> Several probl
2013/7/18 Mark Thomas wrote:
>
> On 18/07/2013 08:00, Violeta Georgieva wrote:
> > Hi,
> >
> > Let's have a lambda expression:
> >
> > x->y->x-y
> >
> > Let's now assign it and invoke it indirectly:
> >
> > f = x->y->x-y; f(2)(1)
>
> Several problems here.
>
> 1. The grammar didn't support function
On 18/07/2013 08:00, Violeta Georgieva wrote:
> Hi,
>
> Let's have a lambda expression:
>
> x->y->x-y
>
> Let's now assign it and invoke it indirectly:
>
> f = x->y->x-y; f(2)(1)
Several problems here.
1. The grammar didn't support functions having multiple sets of parameters.
2. The lambda
Hi,
Let's have a lambda expression:
x->y->x-y
Let's now assign it and invoke it indirectly:
f = x->y->x-y; f(2)(1)
>From the spec we have:
"If the result of evaluating the function name is a LambdaExpression, the
LambdaExpression is invoked with the supplied arguments. If the result of
evalua
