Possible to supress the: "Blocking call to accept-process-output with quit inhibited!!" message?

Question

When running a process from a timer, this message is printed a lot.

Is there a way to suppress the message entirely?

Well the message is there for a reason -- Emacs has been told to wait for output from a process for an *unlimited duration, without allowing the user to interrupt*. Instead of suppressing the message, perhaps it would make best sense to avoid it from being issued in the first place. — phils, Feb 18 '20 at 10:48
Right, this should be worked around somehow, the issue is - I have no idea how. I wanted to run a spell checker on a timer, which seems to be possible to do `wcheck` package does it, however the code is fairly complex and I'd need to spend much more time digging into it to figure out how it works. — ideasman42, Feb 18 '20 at 22:53

score 0 · Answer 1 · answered May 24 '19 at 08:50

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From what I can tell the answer is no.

However, this can be worked around (albeit a hack).

answered May 24 '19 at 08:50

ideasman42

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There is also `inhibit-message` (still a hack, just slightly more concise). – npostavs May 26 '19 at 14:03
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This happens not to work in this case, I think because the message call is from C code. – ideasman42 May 26 '19 at 14:09
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Hmm, I thought the point of `inhibit-message` is that it should work on C code, unlike advising `message` which doesn't. Although looking at its docstring, I notice `inhibit-message` doesn't inhibit logging, just the display. So if you want to keep the log clear too, it's not really that much more concise. – npostavs May 26 '19 at 14:32
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@npostavs `(let ((inhibit-message t) (message-log-max nil)) ...)` is the best way I know of to do both. – phils Feb 18 '20 at 10:40

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