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fix: node.js debugger adds stderr (but exit code is 0) -> shouldn't throw #179
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In this case, maybe concat stderr info too ?
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if it's merged, do I ignore this ?
concat stderr to what ?
to
replacement
?replacement = result.stdout.decode("utf-8").rstrip() + "\n" + result.stderr.decode("utf-8").rstrip()
the order will be changed, and I don't know how to keep original order as it appeared on screen
the original code did not concat stderr
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original code do this in
sys.stderr.write(p_stderr)
Maybe put stderr first.
you can do a new PR.
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the original code is redundant:
it captures stderr,
stderr=subprocess.PIPE,
if stderr is not empty string
if p.wait() != 0 or p_stderr:
it writes it back to stderr
sys.stderr.write(p_stderr)
(it was only captured for condition checking)
the new code:
just don't capture it, stderr is written to stderr anyways
assuming
replacement
(stdout) is also written to stderr,if you don't capture stderr, you don't have to concat it to stderr
I don't see the point of this concatenation, what does it even change ?