bogus error when the RHS of `Table.[]=` is not a compatible type

[alaviss]

Example

import std/tables

var o: Table[int, string]
o[0] = 0

Actual Output

test.nim(4, 2) Error: type mismatch: got <Table[system.int, system.string], int literal(0)>
but expected one of:
proc `[]`(s: string; i: BackwardsIndex): char
  first type mismatch at position: 0
proc `[]`(s: var string; i: BackwardsIndex): var char
  first type mismatch at position: 0
proc `[]`[A, B](t: OrderedTableRef[A, B]; key: A): var B
  first type mismatch at position: 0
proc `[]`[A, B](t: OrderedTable[A, B]; key: A): B
  first type mismatch at position: 0
proc `[]`[A, B](t: TableRef[A, B]; key: A): var B
  first type mismatch at position: 0
proc `[]`[A, B](t: Table[A, B]; key: A): B
  first type mismatch at position: 0
proc `[]`[A, B](t: var OrderedTable[A, B]; key: A): var B
  first type mismatch at position: 0
proc `[]`[A, B](t: var Table[A, B]; key: A): var B
  first type mismatch at position: 0
proc `[]`[A](t: CountTableRef[A]; key: A): int
  first type mismatch at position: 0
proc `[]`[A](t: CountTable[A]; key: A): int
  first type mismatch at position: 0
proc `[]`[I: Ordinal; T](a: T; i: I): T
  first type mismatch at position: 0
proc `[]`[Idx, T; U, V: Ordinal](a: array[Idx, T]; x: HSlice[U, V]): seq[T]
  first type mismatch at position: 0
proc `[]`[Idx, T](a: array[Idx, T]; i: BackwardsIndex): T
  first type mismatch at position: 0
proc `[]`[Idx, T](a: var array[Idx, T]; i: BackwardsIndex): var T
  first type mismatch at position: 0
proc `[]`[T, U: Ordinal](s: string; x: HSlice[T, U]): string
  first type mismatch at position: 0
proc `[]`[T; U, V: Ordinal](s: openArray[T]; x: HSlice[U, V]): seq[T]
  first type mismatch at position: 0
proc `[]`[T](s: openArray[T]; i: BackwardsIndex): T
  first type mismatch at position: 0
proc `[]`[T](s: var openArray[T]; i: BackwardsIndex): var T
  first type mismatch at position: 0
template `[]`(a: WideCStringObj; idx: int): Utf16Char
  first type mismatch at position: 0
template `[]`(s: string; i: int): char
  first type mismatch at position: 0

expression: `[]`(o, 0)

Expected Output

Something that says the last 0 is not a compatible type

[zerbina]

A reduced reproducer:

type Type = object

proc `[]=`(a: Type, b, c: int) =
  discard

var x: Type
x[0] = 1.0

How to report a sensible error, I'm not sure yet (the necessary context to do so is missing), but I don't think [] overloads should be ignored in assignments, which is what's currently happening.

[saem]

Been staring at this for a bit, not sure how we'd improve the error beyond wording or formatting perhaps, but otherwise the core meaning of it would remain intact as we can't find a matching overload. 😬

[alaviss]

For the particular error I posted, this part got me confused:

Error: type mismatch: got <Table[system.int, system.string], int literal(0)>
but expected one of:
proc `[]`[A, B](t: Table[A, B]; key: A): B
  first type mismatch at position: 0
proc `[]`[A, B](t: var Table[A, B]; key: A): var B
  first type mismatch at position: 0

It didn't even occur to me that the RHS was wrong...

[zerbina]

Been staring at this for a bit, not sure how we'd improve the error beyond wording or formatting perhaps, but otherwise the core meaning of it would remain intact as we can't find a matching overload. 😬

What would help is trying to actually try and use a [] overload. While the error says that no [] matched, this doesn't reflect reality.

At the moment, only the []= procedure is tried. If that fails, the compiler simply reports an (incomplete) "not found error":

nimskull/compiler/sem/semexprs.nim

Lines 2219 to 2231 in d393260

	of nkBracketExpr:
	# a[i] = x
	# --> `[]=`(a, i, x)
	a = semSubscript(c, a, {efLValue})
	if a.kind == nkCall:
	case mode
	of noOverloadedSubscript:
	result = bracketNotFoundError(c, a)
	else:
	result = a
	result[0].ident = getIdent(c.cache, "[]=")
	result.add(n[1])
	result = semExprNoType(c, result)

Beyond the incomplete error message, it also means that the following doesn't work:

type Object = object
  a: array[16, int]

proc `[]`(x: var Object, i: int): var int = x.a[i]

var o: Object
o[0] = 1