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fn:identity: make it variadic #1473
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I'm not convinced. What should
fn:identity((1,2), (3,4))return, and why? Whatever it returns, the result doesn't seem to be "identical" to the input.fn:voidis another matter. Since that ignores all its arguments, it might as well ignore a variable number of them.I'm not convinced. What should
fn:identity((1,2), (3,4))return, and why? Whatever it returns, the result doesn't seem to be "identical" to the input.A variadic
fn:identitycan be seen as a generalized variant of Haskell’sconstfunction, which takes two arguments and returns only the first one.fn:identitywould do the same: Only the first argument is evaluted, the remaining ones are ignored.#858 (comment) included an example for this. Here is a simplified version:
$insert-withis called with an function argument that decides how to update the value of a map entry. If we hadidentity#2, we could pass it instead offn($a, $b) { $a }:let $insert-with := fn($map, $key, $value, $update-func) { let $old-value := $map($key) let $new-value := if($old-value) then $update-func($value, $old-value) else $value return map:put($map, $key, $new-value) } let $map := { 'foo': 1 } let $summarize := fn($a, $b) { $a + $b } let $replace := fn($a, $b) { $a } return ( $insert-with($map, 'foo', 2, $summarize)('foo'), $insert-with($map, 'foo', 42, $replace)('foo') )
I'd find this very confusing, identity for me, by definition, is not variadic.
In the use case given, doesn't
fn:identity#1serve equally well?In the use case given, doesn't
fn:identity#1serve equally well?I would assume that
identity#1($value, $old-value)raises an error, but I may be wrong?The function coercion rules say If F has lower arity than T, then F is wrapped in a new function that declares and ignores the additional argument, so supplying
identity#1is equivalent to supplyingfn($x, $y){identity#1($x)}.The function coercion rules say If F has lower arity than T, then F is wrapped in a new function that declares and ignores the additional argument, so supplying
identity#1is equivalent to supplyingfn($x, $y){identity#1($x)}.So I got it right.
fn($a, $b) { $a }is probably still easier to read thanfn($x, $y){identity#1($x)}. Iffn:identitywas variadic, though, we could replacefn($a, $b) { $a }withfn:identity#2.While I could see some use for a variadic
fn:identity, it will be hard to teach how it works (the mere existence offn:identityis already a miracle to almost everyone). Maybe it’s better to drop too esoteric endeavors.The CG agreed to close this issue without further action at meeting 094.
…see #1437 (comment).