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Subject: Re: [Caml-list] Applying labeled function without a label


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On 10/07/2013 17:02, John Carr wrote: 

A function with an unconstrained type variable on the right hand side
of its type can potentially take an unlimited number of arguments.

You have the equivalent of a shift-reduce conflict in a grammar.  The
compiler can choose to reduce (apply f to the argument) or shift (save
the arguments waiting for more).  The choice to shift seems odd but
there may be a situation where it is useful.

If your function were defined

let f ~a = a + 0

it would be monomorphic. The compiler would match an unlabeled argument
with parameter a. 


That is to say : 
# let f ~a = a + 0;; 
val f : a:int -> int = <fun> 
# f 12;; 
- : int = 12 


<blockquote>

<blockquote>
Please, can someone explain the reason behind the following behaviour:


# let f ~a = a;;
val f : a:'a -> 'a = <fun>

if I apply function f, omiting the label, instead of an error I'll get:

# f 12;;
- : a:(int -> 'a) -> 'a = <fun>

... a function that accepts a labeled arguments, that is a function from int
to 'a, and returns a result of this function:

# f 12 ~a:(fun x -> x + 1);;
- : int = 13

and even more, if I apply it to more unlabled arguments:

# f 1 2 3 4 5;;
- : a:(int -> int -> int -> int -> int -> 'a) -> 'a = <fun>

It is very confusing...

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Caml-list mailing list.  Subscription management and archives: https://sympa.inria.fr/sympa/arc/caml-list Beginner's list: http://groups.yahoo.com/group/ocaml_beginners Bug reports: http://caml.inria.fr/bin/caml-bugs 
</blockquote>

</blockquote>


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    <br>
    On 10/07/2013 17:02, John Carr wrote:
    <blockquote cite="mid:201307101502.r6AF2mcZ007014@outgoing.mit.edu">
      <pre>A function with an unconstrained type variable on the right hand side
of its type can potentially take an unlimited number of arguments.

You have the equivalent of a shift-reduce conflict in a grammar.  The
compiler can choose to reduce (apply f to the argument) or shift (save
the arguments waiting for more).  The choice to shift seems odd but
there may be a situation where it is useful.

If your function were defined

let f ~a = a + 0

it would be monomorphic. The compiler would match an unlabeled argument
with parameter a.</pre>
    </blockquote>
    <br>
    That is to say :<br>
    # let f ~a = a + 0;;<br>
    val f : a:int -&gt; int = &lt;fun&gt;<br>
    # f 12;;<br>
    - : int = 12<br>
    <br>
    <blockquote cite="mid:201307101502.r6AF2mcZ007014@outgoing.mit.edu">
      <pre>
</pre>
      <blockquote>
        <pre>
Please, can someone explain the reason behind the following behaviour:


# let f ~a = a;;
val f : a:'a -&gt; 'a = &lt;fun&gt;

if I apply function f, omiting the label, instead of an error I'll get:

# f 12;;
- : a:(int -&gt; 'a) -&gt; 'a = &lt;fun&gt;

... a function that accepts a labeled arguments, that is a function from int
to 'a, and returns a result of this function:

# f 12 ~a:(fun x -&gt; x + 1);;
- : int = 13

and even more, if I apply it to more unlabled arguments:

# f 1 2 3 4 5;;
- : a:(int -&gt; int -&gt; int -&gt; int -&gt; int -&gt; 'a) -&gt; 'a = &lt;fun&gt;

It is very confusing...

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</pre>
      </blockquote>
      <pre></pre>
    </blockquote>
    <br>
  

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