Wrong match case selected

Compiler version

3.2.0

Minimized code

class Baz:
  inline def foo: Int = bar(zero)
  inline def bar(inline i: Int): Int = inline i match
    case 0 => 0
    case _ => 1
  private inline def zero = 0

println(Baz().foo)

Output

1

Expectation

0

See also #15893

[dwijnand]

It’s a little confusing to me that “disjointness” is a factor in match type reduction but not in inline matches. Makes it more work to remember the different semantics. I’m guessing it’s required or perhaps much preferred.

[Bersier]

@odersky thank you for clarifying.

A few more observations:

1. This behavior (and how it differs from match type reduction) doesn’t currently seem to be fully documented.

2. One might naively assume the inline match to be equivalent to inline if i == 0 then 0 else 1 (but the latter leads to a compile-time error).

3. In both cases, it also seems at odds with the part of the documentation that says It guarantees that inlining actually happens instead of being best-effort. According to that statement, one might naively assume code reduction to proceed somewhat as follows:

class Baz:
  inline def foo: Int = bar(zero)
  inline def bar(inline i: Int): Int = inline if i == 0 then 0 else 1
  private inline def zero = 0

println(Baz().foo)

--> (because zero is guaranteed to be inlined)

class Baz:
  inline def foo: Int = bar(0)
  inline def bar(inline i: Int): Int = inline if i == 0 then 0 else 1

println(Baz().foo)

--> (because bar is guaranteed to be inlined)

class Baz:
  inline def foo: Int = inline if 0 == 0 then 0 else 1

println(Baz().foo)

--> (inline if reduction)

class Baz:
  inline def foo: Int = 0

println(Baz().foo)

--> (because foo is guaranteed to be inlined)

println(0)