It should because:

    $f = list($c, $d) = [1, 2];

does exactly as you describe and this is literally just a shorthand syntax for accessing list().

Thanks. What about:

  $c = 2;
  $d = 3;

  $f = [$c, $d] = [1, 2];

Is $f: 2, 3 or 1, 2? I assume 1, 2 because if not then it's both an array constructor and destructurer at the same time which would be interesting.

Plus:

  $f = $c = $d;

is 3;

Let's break this up a bit, remembering that assignment is right-associative:

  $c = 2; # Assign 2 to $c
  $d = 3; # Assign 3 to $d

  [$c, $d] = [1, 2]; # Assign 1 to $c, assign 2 to $d

  $f = [$c, $d]; # Assign [1, 2] to $f