chg: populate CHGHG if not set
Normally, chg determines which hg
executable to use by first consulting the
$CHGHG
and $HG
environment variables, and if neither are present defaults
to the hg
found in the user's $PATH
. If built with the HGPATHREL
compiler
flag, chg will instead assume that there exists an hg
executable in the same
directory as the chg
binary and attempt to use that.
This can cause problems in situations where there are multiple actively-used
Mercurial installations on the same system. When a chg
client connects to a
running command server, the server process performs some basic validation to
determine whether a new command server needs to be spawned. These checks include
things like checking certain "sensitive" environment variables and config
sections, as well as checking whether the mtime of the extensions, hg's
__version__.py
module, and the Python interpreter have changed.
Crucially, the command server doesn't explicitly check whether the executable it
is running from matches the executable that the chg
client would have
otherwise invoked had there been no existing command server process. Without
HGPATHREL
, this still gets implicitly checked during the validation step,
because the only way to specify an alternate hg executable (apart from $PATH
)
is via the $CHGHG
and $HG
environment variables, both of which are checked.
With HGPATHREL
, however, the command server has no way of knowing which hg
executable the client would have run. This means that a client located at
/version_B/bin/chg
will happily connect to a command server running
/version_A/bin/hg
instead of /version_B/bin/hg
as expected. A simple
solution is to have the client set $CHGHG
itself, which then allows the
command server's environment validation to work as intended.
I have tested this manually using two locally built hg installations and it
seems to work with no ill effects. That said, I'm not sure how to write an
automated test for this since the chg
available to the tests isn't even built
with the HGPATHREL
compiler flag to begin with.