Error looked something like:
```
% echo 'f(zpty):8: unmatched '
_zsh_autosuggest_capture_completion:zpty:9: no such pty command: zsh_autosuggest_completion_pty
_zsh_autosuggest_capture_completion:zpty:14: no such pty command: zsh_autosuggest_completion_pty
_zsh_autosuggest_capture_completion:zpty:21: no such pty command: zsh_autosuggest_completion_pty
```
According to `man zshmodules`, the args to `zpty` are "concatenated with
spaces between, then executed as a command, as if passed to the eval
builtin." So we need to escape the `$` so that `$1` is passed to eval
instead of the value of `$1`.
The `zsh -f` running in the PTY doesn't know about the non-exported
variables and functions defined in the original shell, thus can't make
suggestions for them. Run local functions in the PTY instead of a new
`zsh` process.
We have to handle things differently based on whether zle is active or
not (async vs. sync mode).
For unknown reasons, the pty will occasionally quit running. In these
cases, we still want to remove it so that a fresh one can be created. We
don't actually need this check because error messages from `zle` and
`zpty` are redirected to /dev/null.
One sure way to kill all currently running pty's is to run `exit` in a
subshell. Even without zsh-autosuggestions loaded, the following works:
% zmodload zsh/zpty
% zpty -b foo cat
% zpty -b bar cat
% zpty
(31689) bar: cat
(31666) foo: cat
% $(exit)
% zpty
(finished) bar: cat
(finished) foo: cat
Specific case where this matters is following:
Be in vi insert mode with some text in the buffer and the cursor at the
end of the buffer. Press `esc` to trigger `vi-cmd-mode widget`, then
before the cursor moves (KEYTIMEOUT), press `h` to trigger
`vi-backward-char` widget. When `vi-cmd-mode` original widget exits,
KEYS_QUEUED_COUNT will be non-zero and the suggestion will be lost.
Eric Freese
Martin Väth
Harm te Hennepe
Benjamin Denhartog
babaorum