outplace argument _

[timotheecour]

proposal

Support the outplace argument placeholder _ to convert an inplace algorithm into an outplace one for the common case where the outplace argument is assumed to be initialized to its default value for its (inferred) type.

Example

proc merge[T](result: var seq[T], x, y: openArray[T]) = ...
let x = [1,3]
let y = [2,5]
let a = merge(_, x, y)
let a2 = _.merge(x,y) # also works

# same as:
var a: seq[int]
merge(a, x, y)

# also same as:
import std/sugar
let a3 = seq[int].default.dup(merge(x, y))
  # less readable and not DRY since seq[int] must be specified (can't do type inference here)

Notes

• the outplace argument can appear in any position, not just the 1st, as its position is given explicitly by _:

proc fn[T](a: seq[T], b: var T) = ...
let a = fn(@[1,2], _) # works
let a2 = @[1,2].fn(_) # also works

• sugar.dup is still useful in the cases where you need a non-default value for the outplace argument, or when type inference wouldn't be possible, because the generic is only inferred from the outplace argument instead of other arguments.

implementation

• requires compiler support.
• in sigmatch, the arguments are implicitly reordered so that _ is matched last, at which point all generic parameters must be bound, or a sigmatch error occurs.

links

• Question about dup syntax - Nim forum https://forum.nim-lang.org/t/7240#45775
• [TODO] [WIP] new outplace operator without intermediate copies/temporaries by timotheecour · Pull Request #13337 · nim-lang/Nim

[metagn]

I don't mean to delay or complicate anything, but thoughts on this?

proc merge[T](result: var seq[T], x, y: openArray[T]) = ...
let x = [1, 3]
let y = [2, 5]
merge(let a, x, y)
(let a2).merge(x,y) # torn on this one

proc fn[T](a: seq[T], b: var T) = ...
fn(@[1, 2], let a) 
@[1, 2].fn(let a2)

I believe some other language does something like this but I forget

[github-actions]

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