brogoff@speakeasy.net wrote:
> On Sun, 22 Sep 2002, Alessandro Baretta wrote:
> 
>>Let me suggest another idea where you might better spend 
>>your time. I can write
>> > let f x : 'a -> 'a = x
> 
> 
> Does this do what you think it does? What is the type of f?
> 
> Incidentally, I'd prefer some form of separate, forward type declaration.
> 
> -- Brian
> 
> 

Nope. Right, the colon-type-equals construct is for the 
definition of the return type, as opposed the actual type of 
f. Let me then rework my suggestion as

method f (x:'a) : 'a.'a = x

which is now actually equivalent to

let f (x:'a) : 'a = x

Alex

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