使用flock时是否需要锁定文件是全局的

Question

I am setting up cron job for single instance. Does the lock below need to be global? I believe so, in order that it stays in scope until the program finishes. Or at least outside of try/except block. Also, return values should be positive from Python? Seems -2 returns 254 on echo $? in bash.

import time, fcntl, sys

LOCK_FILE = '/tmp/test_flock.lock'
lock = None    
def do_wait():
    print ('waiting N sec')
    time.sleep(3)

def main(argv=None):
    try:
        global lock        
        lock = open(LOCK_FILE,'w')
        fcntl.flock(lock, fcntl.LOCK_EX | fcntl.LOCK_NB)
        print ("got flock")
    except IOError as err:
        print ("Could not obtain lock file")
        return -2

    if argv is None:
        argv = sys.argv

    try:
        print ('entering main')
        print ('waiting')
#        raise ValueError("Error raised")
        do_wait()
        print ('done')
    except Exception as err:
        print ("Exception in main")
        return -1

if __name__ == '__main__':
    sys.exit(main())

Answer 1

Seems -2 returns 254 on echo $? in bash.

The Bash is interpreting the -2 as 254 probably due to how it handles negative numbers which I believe is by returning 8-bit unsigned integers.

Soln: Use a postive integer number > 0 but < 256. Bash will see a return code > 0 as an error.

Does the lock below need to be global? I believe so, in order that it stays in scope until the program finishes.

Ans: If there are no other conflicts with locking the file than I would say that the global lock is fine.