The code is like this:
os.execlp('python', 'python', 'child.py', #other args#) # this works
os.execlp('python', 'child.py', #other args#) # this doesn't work
I read this question: execlp() in python
But I'm still confused. The answer said:
The first argument is the program to execute (found on the PATH). The
rest are the sys.argv arguments to the program.
However, if I run: python child.py 1 2 3
and the sys.argv
of this process would be ["child.py", "1", "2", "3"]
, where the python
doesn't exist. Then why should I add python
as the second parameter of os.execlp
?
When python is executed, it creates sys.argv
for you. The values in that list are based on the arguments passed to it by the operating system, but it leaves off the sys.executable
value from that list.
In other words, when Python is invoked, it sets sys.argv
to everything but it's own executable.
When you invoke a new executable via os.execlp()
, you still need to include Python in that as that is what executable that the OS will run. The first two values of what you a pass to os.execlp()
are still required, whatever you find in sys.argv
later on.
The second python is a name for python, it can be any string, but it has to be there.
See the second paragrph of http://docs.python.org/3/library/os.html?highlight=os.exec#process-management