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A to =>A automatic conversion
Wed, 2009-09-23, 23:24
Hi all,
I apologize in advance if I ask a question that has already been asked
(but I did not find the answer in the ML archives) and if my vocabulary
is not the correct one.
If I understand correctly, scala will automatically convert expressions
of type (=>A) to A (and the opposite way) as needed.
But, if won't convert expressions of type (A=>B) to ((=>A)=>B) and
neither ((=>A)=>B) to (A=>B). Is there a way to allow this, knowing that
the conversion functions (using implicit conversion of (=>A) to A) are:
def f1[A,B](f:(=>A)=>B):A=>B = f(_)
def f2[A,B](f:A=>B):((=>A)=>B) = f(_)
And is there a reason for those type of implicit conversions not to be
implemented?
Thu, 2009-09-24, 13:47
#2
Re: A to =>A automatic conversion
> The notation (=> A) refers to a call-by-name parameter of type A; it
> is not a
> type in and of itself. For instance, it is not legal to write:
>
> val a: => A = ...
>
> This recent thread has more info on the topic:
> http://www.nabble.com/
> Re%3A--scala--question-about-function-types-p25473653.html
Thanks for the clarification. I was thinking =>A was ()=>A.
But knowing know that (=> A) is a call-by-name parameter, is there a
reason for scala not to have implicit conversions from f(A) to f(=>A) and
f(=>A) to f(A)? I mean if I have:
val foo: ((=>A)=>B)=>Unit = ...
and
val bar: (A)=>B = ...
what would be the problem to allow to call foo(bar)?
Thu, 2009-09-24, 13:47
#3
Re: A to =>A automatic conversion
Hmm, I didn't realise val foo: ((=>A)=>B)=>Unit would be valid syntax.
scala> val foo: (=>String)=>Unit = null
foo: (=> String) => Unit = null
scala> val foo: (=>String) = null
error: no by-name parameter type allowed here
val foo: (=>String) = null
Please could someone explain this?
2009/9/24 Benjamin Lerman :
>> The notation (=> A) refers to a call-by-name parameter of type A; it
>> is not a
>> type in and of itself. For instance, it is not legal to write:
>>
>> val a: => A = ...
>>
>> This recent thread has more info on the topic:
>> http://www.nabble.com/
>> Re%3A--scala--question-about-function-types-p25473653.html
>
> Thanks for the clarification. I was thinking =>A was ()=>A.
>
> But knowing know that (=> A) is a call-by-name parameter, is there a
> reason for scala not to have implicit conversions from f(A) to f(=>A) and
> f(=>A) to f(A)? I mean if I have:
>
> val foo: ((=>A)=>B)=>Unit = ...
>
> and
>
> val bar: (A)=>B = ...
>
> what would be the problem to allow to call foo(bar)?
>
> --
> Benjamin
>
Thu, 2009-09-24, 14:07
#4
Re: A to =>A automatic conversion
FWIW, I hit this problem about two years ago but have found no solution.
Stream's foldRight method should be using call-by-name parameter types
in the function argument.
Benjamin Lerman wrote:
>> The notation (=> A) refers to a call-by-name parameter of type A; it
>> is not a
>> type in and of itself. For instance, it is not legal to write:
>>
>> val a: => A = ...
>>
>> This recent thread has more info on the topic:
>> http://www.nabble.com/
>> Re%3A--scala--question-about-function-types-p25473653.html
>>
>
> Thanks for the clarification. I was thinking =>A was ()=>A.
>
> But knowing know that (=> A) is a call-by-name parameter, is there a
> reason for scala not to have implicit conversions from f(A) to f(=>A) and
> f(=>A) to f(A)? I mean if I have:
>
> val foo: ((=>A)=>B)=>Unit = ...
>
> and
>
> val bar: (A)=>B = ...
>
> what would be the problem to allow to call foo(bar)?
>
>
Thu, 2009-09-24, 14:17
#5
Re: A to =>A automatic conversion
Tony Morris a écrit :
> FWIW, I hit this problem about two years ago but have found no solution.
> Stream's foldRight method should be using call-by-name parameter types
> in the function argument.
That's *exactly* the reason I'm asking this question.
Thu, 2009-09-24, 22:57
#6
Re: A to =>A automatic conversion
I fixed foldRight myself (by generalising the type constructor) and put
up with the minor subsequent nuances.
Benjamin Lerman wrote:
> Tony Morris a écrit :
>
>> FWIW, I hit this problem about two years ago but have found no solution.
>> Stream's foldRight method should be using call-by-name parameter types
>> in the function argument.
>>
>
> That's *exactly* the reason I'm asking this question.
>
>
val a: => A = ...
This recent thread has more info on the topic:
http://www.nabble.com/Re%3A--scala--question-about-function-types-p25473653.html
- Colin