Hope this helps,
Exactly what I wanted to know! Thanks!
Dawid
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Thanks for your answer; now I'll try to sate my point in a clear way,
since you misunderstood all my questions.
I put 'a in the interface:
val fu : 'a - 'a
and int in the implementation:
let fu x = x + 1
this interface doesn't reflect the implementation, ocaml will reject it in any
way
OK,
Yes you are right my first answer was totally out of scope,
I will try to make it forgive and forget with this new one :)
Still take it with care while it seems you are far more knowledged than me
to answer to your own question ;-)
Just an elementary question.
I put 'a in the interface:
val
Just an elementary question.
I put 'a in the interface:
val fu : 'a - 'a
and int in the implementation:
let fu x = x + 1
So I have universal quantification: for any type 'a function fu can
consume the argument. So my implementation doesn't comply with that
universal quatification. And the
