Use @ instead of % as % is already an operator

[Mike96Angelo]

Since % is the modulus operator I think it we be better to use a different symbol, perhaps @

let even = 8 |> @ % 2 === 0;  // 8 % 2 === 0

In some other programming languages that have the pipe operator they support piping is both directions:

let a = 3
  |> square(@)
// let a = square(3)
  
let b = square(@) 
  <| 2
// let b = square(2)

There is some confusion if both are use in the same expression:

let a = 1 |> add(@, @) <| 2
// let a = add(1, 2) or is it  add(2, 1)

to get over this confusion we can add the corresponding < or > to the @

let a = 1 |> add(@>, <@) <| 2
// let a = add(1, 2) no more confusion 

let a = 1 |> add(<@, @>) <| 2
// let a = add(2, 1) no more confusion 

[ljharb]

How does that interact with decorators?

[Mike96Angelo]

Since decorators require two arguments @<expression that results in a decorator function> <thing to decorate> we can determine that an @ that receives no arguments that follows a |> can be treated as an injection point instead of a decorator.

if we allow directional piping and use directional injection points (@>, <@) there is no conflict with decorators at all.

[MeguminSama]

@ has been excluded as a topic reference, noted here:
#232 (comment)
#91 (comment)

[bogdanbiv]

Please close this issue, as it was cleared in the previous comment.