Glynn Clements writes:
Although, depending upon the OS, setting SIGCHLD to
SIG_IGN may cause processes to be reaped automatically
(i.e. not become zombies), so that's a possible
alternative.
I think I've got it under control now. I'm using this
wrapper to make sure there are no
David Brown wrote:
Both [waitForProcess and getProcessExitCode] will throw
an exception if the process terminated on a signal.
So if I terminate a process manually, I'll have to wait for
the ExitCode to avoid a zombie process, and waiting for the
ExitCode invariably throws an
On Thu, Oct 28, 2004 at 06:27:42AM +0200, Peter Simons wrote:
Glynn Clements writes:
Both [waitForProcess and getProcessExitCode] will throw
an exception if the process terminated on a signal.
So if I terminate a process manually, I'll have to wait for
the ExitCode to avoid a zombie
Peter Simons wrote:
Both [waitForProcess and getProcessExitCode] will throw
an exception if the process terminated on a signal.
So if I terminate a process manually, I'll have to wait for
the ExitCode to avoid a zombie process, and waiting for the
ExitCode invariably throws an
John Goerzen writes:
Assuming it is based on wait() or one of its derivatives,
and I suspect it is, you cannot call it more than once
for a single process.
That's what I _assume_, too, but a definite answer would be
nice.
In the meanwhile, I have found out that it might not be safe
to
Peter Simons wrote:
John Goerzen writes:
Assuming it is based on wait() or one of its derivatives,
and I suspect it is, you cannot call it more than once
for a single process.
That's what I _assume_, too, but a definite answer would be
nice.
In the meanwhile, I have found out
What will happen if I call getProcessExitCode for the same
process twice? Will that block? Cause an error? Or return
the same child's exit code again?
I assume the function is (under Unix) based on wait(2),
right? In that case, how does the following warning from the
manual page translate to
On 2004-10-26, Peter Simons [EMAIL PROTECTED] wrote:
What will happen if I call getProcessExitCode for the same
process twice? Will that block? Cause an error? Or return
the same child's exit code again?
Assuming it is based on wait() or one of its derivatives, and I suspect
it is, you cannot
