why are pipeline commands (allowed to be) executed in subshells?

[Philippe Cerfon]

Hey.

I noticed that bash (an other shells) seem to execute every command of
a pipeline:
[!] cmd1 [ | cmd2 …]
in separate subshell, but only when there's actually at least one pipe
operator ([!] cmd1 alone is, as per POSIX, also a pipeline).

For example:
    n=1
    echo foo | { n=2 ;}
    echo $n
yields:
    1

Or:
    n=1
    echo foo | { n=2 ;}
    echo $n
yields:
    1

And:
    n=1
    { n=2 ;} | echo $n
    echo $n
yields:
    1
    1
meaning that there is one subshell per command of the pipeline, and
not one for all.

It's clear that any non-built-in command/programs are executed not in
the same shell environment, but why are all shell commands?

Any what in POSIX even allows this?

https://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html:

- There is no command substitution involved.
- What's said in 2.9.1 Simple Commands about subshells is only about
redirections, when there is no command name.
- 2.9.2 Pipelines says "If the pipeline begins with the reserved word
! and command1 is a subshell command…", but the next sentence clearly
refers to the ( ) grouping command.
- there's no asynchronous list, which would make the commands executed
in subshells

And if the argument were "well it doesn't mandate it but doesn't
forbid it either"; then why is the one element pipeline:
   [!] command
e.g. n=2
not executed in a subshell either?
If POSIX would allow that behaviour for pipelines, wouldn't that kill
any assignments?

Or why are AND/OR lists not:
   n=1
   echo foo && { n=2 ;}
   echo $n
yields:
   foo
   2
and:
   n=1
   { n=2; } && { n=$((n+1)) ; echo $n; n=5; }
   echo $n
yields:
   3
   5


Regards,
Philippe