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Anonymous Function Definition
f#book

Ok, another noob question... I've noticed that when you look at the definition of a function that has an anonymous function as a parameter it uses the '->' notation to separate parameter and return types. For example:

let doSomething p1 p2 p3 = p2 p1 p3

Is defined as:

val doSomething : 'a -> ('a -> 'b -> 'c) -> 'b -> 'c

Which makes sense. But if you show the definition of Seq.unfold it looks like this:

val it : (('a -> ('b * 'a) option) -> 'a -> seq<'b>) = <fun:clo@0-2>

First, I'm not understanding why there isn't a '->' between the tuple ('b * 'a) and option, like this:

val it : (('a -> ('b * 'a) -> option) -> 'a -> seq<'b>) = <fun:clo@0-2>

Also, why is there that funky = <fun:clo@0-2> at the end and the definition surrounded by braces? This threw me off a little as I was trying to understand how to use unfold.

I tried looking around for an answer but came up empty handed. I may have read about it in the Expert F# book but I cant seem to remember where.

TIA!

m

.

Seq.unfold's first argument is a function which itself takes items of type 'a and returns an optional tuple. In this case, ('b*'a) is the type parameter for the option type (that is, the result of the function could be None or Some p, where p is a pair of a 'b and an 'a). This is made more confusing by the fact that generic types can take their parameters .NET-style, like option<'b*'a>, or ML-style, like ('b*'a) option, or byte array, or int list, etc.

As to your second question, the = sign is not part of the type; instead, F# is telling you that the automatic 'it' variable has type (('a -> ...), and is equal to whatever the rest of the line said, but which you have elided. For me, the remainder of the line finishes with <fun:clo@0>, which is F#'s way of displaying a compiled closure. Compare this to what happens if you type in a literal, such as [1;2;3]; in the case of lists the printed representation of the value is more useful.

By on 8/14/2009 10:37 AM ()

Ah, ok, that makes perfect sense now on the option definition, I don't know what I was thinking. My brain is a little mushy right now from all this... :)

I see what your saying on the closure as well. If I keep redefining that same method over and over the function name suffix increments ("clo{at}0-1", "clo{at}0-2", ...) as it should.

Thanks for your help!

m

By on 8/14/2009 10:51 AM ()

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