Getting Error :init: Expected printf output from shell

[didibus]

When launching spacemacs which uses exec-path-from-shell I get the following error from it:

Error (use-package): exec-path-from-shell :init: Expected printf output from shell, but got: "%  

�[01;36muser@cloud] �[00m"

I'm sure it's due to some of the things I do or source from my .zshrc, but I'm not sure what, and I'd rather exec-path-from-shell could handle the customization of my zsh, then me having to get rid of it.

Anyone got this problem before?

P.S.: Same thing happens if I manually run (exec-path-from-shell-initialize) or (exec-path-from-shell-copy-env) for any variable.

[purcell]

Yeah, your shell is clearly printing stuff out when it runs commands. It looks like a prompt, so that's a bit strange to me, because shells don't normally print out prompts when invoked with -c to run a command. In other words, your config is a bit funky. :-)

You have a couple of options: either fix the spurious prompt printing (and as a shortcut for that, you can try skipping some config if $INSIDE_EMACS is set), or change exec-path-from-shell-arguments to remove the -i flag so that .zshrc is not sourced at startup when exec-path-from-shell runs zsh.

I also use zsh and I don't have this problem. I put all my env var setup code in .zshenv, and then only have a few aliases and prompt config things in .zshrc. That setup works for me with or without the -i flag.

[didibus]

I figured out this is caused by exporting a custom PROMPT in my zshrc file, like this:

export PROMPT="%{$fg_bold[cyan]%}%n@cloud] %{$reset_color%}%";

But if I run /bin/zsh -l -i -c "printf \"Path: %s\" $PATH" it does not print out the prompt. So I'm not sure why it does from exec-path-from-shell.

[purcell]

@didibus try setting exec-path-from-shell-debug to t so you can confirm what exec-path-from-shell is receiving from your shell.

[didibus]

I get:

Invoking shell /bin/zsh with args ("-l" "-i" "-c" "/usr/bin/printf '__RESULT\\000%s\\000%s\\000__RESULT' \"${PATH-851f8100b2783cafbcce2d6e0476fded}\" \"${MANPATH-851f8100b2783cafbcce2d6e0476fded}\"")
Shell printed: "%  
 
�[01;36mdidibus@cloud] �[00m"

And:

Error (use-package): exec-path-from-shell/:init: Expected printf output from shell, but got: "%  
 
�[01;36mdidibus@cloud] �[00m"

I can't seem to find what in my zshrc changes the prompt to return this. Setting the prompt explicitly at the end to "" still doesn't work, I get:

Invoking shell /bin/zsh with args ("-l" "-i" "-c" "/usr/bin/printf '__RESULT\\000%s\\000%s\\000__RESULT' \"${PATH-d0f2de2d4ac0b7a85a8b54f3db589d51}\" \"${MANPATH-d0f2de2d4ac0b7a85a8b54f3db589d51}\"")
Shell printed: "%  
 
"

instead.

Removing the -i flag works.

Also it looks like INSIDE_EMACS is not set, since I tried putting if [ -z "${INSIDE_EMACS+1}" ]; then around my entire zshrc file, and it did not work.

[purcell]

Based on the above, I suspect there's a mistake in one of your shell start-up files, or you're misunderstanding which of those files are evaluated by zsh when -i is either set or unset.

(INSIDE_EMACS is set by Emacs itself when running interactive shells, e.g. via M-x shell. It's not surprising to me that shell-command-to-string does not set it.)