Re: process substitution flawed by design

[Greg Wooledge]

On Tue, Feb 21, 2017 at 02:18:03PM +0100, Florian Mayer wrote:
> for mutex --lock I use a tool which I wrote myself.

What does it do?

> The following code assumes the lock to be in state not-taken before the 
> snippet runs.

What lock?

> echo foo  | tee \
>     >(mutex --lock; echo before; cat; echo after; mutex --unlock) \
>     >(mutex --lock; echo foobar; mutex --unlock) \
>     > /dev/null | cat

> However, the code occasionally just deadlocks.
> I already found out that deadlocks only occur if I try to read from 
> stdin in one of the two >()-blocks.
> 
> How could I try to debug this?

I can't imagine how one would debug your mutex program without knowing
anything about it.

> The only reason I can think of is that somehow cat never exits.

Which cat?  You've got two.  I don't even know what the second one is
for, except to make the problem more complicated.  You're not feeding
it any input.