The shell forks, and the child process tries to
execute child.sh using
execve(). If this succeeds, the arguments you see in
ptree show the
child script. If it fails (because there is no #!
line), the child
process interprets the script, but the process args
still match those of
the
Henrik henrik.lyngga...@gmail.com wrote:
Why doesn't the child process change the process name after the failed
execve() ? is it just history or is this a feature ?
The process name cannot be changed. THe only way is to create a new process
with a new name.
You can change argv[] for a
Hi
I have seen a weird behaviour from bash both under Solaris and Linux, most
likely it is a feature but I can't seem to find the answer with google :-\
I have two scripts
test.sh:
#!/usr/bin/bash
echo parent
./child.sh
sleep 120
child.sh:
echo child
sleep 120
If I run the two scripts like
test.sh:
#!/usr/bin/bash
echo parent
./child.sh
sleep 120
child.sh:
echo child
sleep 120
If you omit the #!/bin/bash from child.sh, I suspect the original bash will
look at the file, recognize it's executable, but since the file doesn't
specify a new command interpreter, it's
If you omit the #!/bin/bash from child.sh, I suspect
the original bash will
look at the file, recognize it's executable, but
since the file doesn't
specify a new command interpreter, it's equivalent to
. ./child.sh meaning
the parent script simply sources the child script.
Not in a new
Henrik wrote:
If you omit the #!/bin/bash from child.sh, I suspect
the original bash will
look at the file, recognize it's executable, but
since the file doesn't
specify a new command interpreter, it's equivalent to
. ./child.sh meaning
the parent script simply sources the child script.
Not in
