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is there some way to debug a process ignoring child processes that tries to connect to the debugger? #649
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Node.js v8.1.0 will let each process bind to a separate random port with |
thnks for answering! and yes i think that at least it will let the application to run normally without exits of child processes. just one concern, in the theorical case that some package that uses a child process gets published with a |
No, In fact, if you wanted the option to break at an early |
clear now, looking forward for random ports thnk you! |
i'm trying to debug my node.js application (using
--inspect
), the application uses some third-party modules that make use of thechild_process
module internally.when i try to debug my app and when some part of the code reach other part that uses
child_process
a message like this is printedStarting inspector on 0.0.0.0:9229 failed: address already in use
and the child process exits with an exit code of12
(which is documented in docs), what i understand is that all nodejs child processes by default tries to connect to the inspector on the same port (9229
by default), however i'm trying to find a way to tell node that all child processes should not try to connect to the inspector, i'm only interested in debugging all the code that is running in the same process, this is something that was possible using the now deprecated node-inspector package, withnode-inspector
all code can work normally because child processes does not try to connect to the debugging session, which makes debugging all the code in the main process possible.is there something i can do to achieve this? it is really annoying that the application does not work correctly in debugging mode for this reason. thanks!
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