Why does some text disappear when using a quoted command substitution in this expression?

[lacygoill]

Consider this snippet:

set x a.b.c
echo "$x$(printf d.e.f)"

It outputs this:

.e.f

But I don't understand the result. I would have expected this instead:

a.b.cd.e.f

Just like in any of these similar snippets:

set x a.b.c
echo $x"$(printf d.e.f)"

a.b.cd.e.f

set x a.b.c
echo "$x"$(printf d.e.f)

a.b.cd.e.f

set x a.b.c
echo $x$(printf d.e.f)

a.b.cd.e.f

set x a.b.c
set y "$(printf d.e.f)"
echo "$x$y"

a.b.cd.e.f

The only syntax that I know of which might explain why some text disappears, is the cartesian product documented at man fish-language, section PARAMETER EXPANSION, subsection Combining lists (Cartesian Product).

For example:

set x a b c
set --erase y
echo $x$y

But this requires one of the operands to evaluate to an empty list. And in the original snippet, none of the operands evaluates to an empty list:

set x a.b.c
set y $(printf d.e.f)
set --show x
set --show y

$x: set in global scope, unexported, with 1 elements
$x[1]: |a.b.c|
$y: set in global scope, unexported, with 1 elements
$y[1]: |d.e.f|

Adding quotes around the command substitution in the y assignment does not change the results.

Where can I learn more about the syntax which explains why a.b.cd disappears?

[krobelus]

Oof that's embarassing. The behavior is wrong, we accidentally add the results of the command substitution to the variable name. Fixed by 1b668f5, thanks for reporting