Operators in $if

[scan]

Another issue I have found, not documented, is using an operator in an $if, like so:

$if (length list) > 2
$if isValid (a+b)

And other examples (with ==) and such... they all produce the error 'Unexpected >' or 'Unexpected =' and so on. Is that intended? If I define a function

let eq = (==)

It works fine, and operators are just functions, so shouldn't that work?

Addition:

$if (length list) `eq` 1

yields the same problem.

[snoyberg]

No, you can't use an operator like that, that's why it's not
documented ;). Doing so would require knowing all of the precedence
levels, which is not available to the TH code involved. You can
instead do something like this:

$if (>) (length list) 2
$if isValid $ (+) a b

On Tue, Mar 15, 2011 at 8:36 PM, scan
[email protected]
wrote:

Another issue I have found, not documented, is using an operator in an $if, like so:

   $if (length list) > 2
   $if isValid (a+b)

And other examples (with ==) and such... they all produce the error 'Unexpected >' or 'Unexpected =' and so on. Is that intended? If I define a function

   let eq = (==)

It works fine, and operators are just functions, so shouldn't that work?

Reply to this email directly or view it on GitHub:
https://github.com/snoyberg/hamlet/issues/13

[scan]

Is that a general restriction or can that be solved?

[snoyberg]

I don't think it's solvable, but more to the point, I don't want to solve it. I think allowing arbitrary infix syntax is a bad direction for Hamlet. I agree 100% that it would be useful in many situations, but it also leads to some ambiguity. For example, the same Hamlet template could have different meanings depending on which modules are in scope when it is called.

On the other hand, we could simply force all operators to have the same precedence rules, but then it won't line up with the Haskell code.

[scan]

Yes, ok, thanks. It makes sense there, after all, it makes no difference for the program, only to the programmer.