let-in if do-expr is problematic? (was: Re: proposal: let in if parentheses)

[Herbert Vojčík]

Hi!

It would be nice to know if do expressions have some a chance, otherwise 
some other syntax for let-in would be really helpful, especially now 
that we have arrow functions.

I would propose to use different variant of let (maybe also const):

OP 1:

   let in a = b(), if (a) a.c();

OP 2:

   let in a = b(), if (a) c(a);

Instead of
   const big = raw => {
     let cooked = cook(raw);
     return consumer => {
       // do things with consumer and cooked
     };
   };

   const big = raw =>
     let in cooked = cook(raw), consume => {
       // do things with consumer and cooked
     };

In short,

   let in binding = expr, stmt|expr

It may work for `const in` as well.

Herby

P.S.: Alternative syntax is "let a=3, b=4, ..., in foo(a,b,c,d)" but 
this can only tell late if it is plain let-up-to-end-of-scope or 
local-scope-let, so not sure if that may be a problem; OTOH you can 
chain more of them and resembles classical let-in better.

Isiah Meadows wrote on 21. 8. 2018 20:17:
> It's possible, but the ability to optionally destructure is what would 
> make this feature worth it - I feel this should wait for pattern 
> matching to be added first, though.
> 
> On Tue, Aug 21, 2018 at 11:10 Jordan Harband <ljharb at gmail.com 
> <mailto:ljharb at gmail.com>> wrote:
> 
>     ```
>     {
>        let a = b();
>        if (a) {
>          c(a);
>        }
>     }
>     ```
> 
>     On Tue, Aug 21, 2018 at 10:54 AM, Ali Rahbari <rahbari at gmail.com
>     <mailto:rahbari at gmail.com>> wrote:
> 
>         while it's possible to use let keyword in for loop parentheses,
>         it's not possible to use it in if parentheses.
> 
>         There are two use cases for this:
> 
>         *1- if (let a = b()) a.c();*
>           this can be done using optional chaining which is proposed:
>         b()?.c();
> 
>         *2- if (let a = b()) c(a);*
>         this or more sophisticated patterns can't be done in any way
>         other than this:
>         let a = b();
>         if (a) c(a);
> 
>         the problem here beside more line of codes, is *a *is defined
>         outside of if scope.
> 
> 
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