Question

import sys  

def checkarg():  
    try:  
        filename=str(sys.argv[1])  
        if filename=="-mycommand":  
            print "SPECIFIC_TEXT"  
            sys.exit()  
        else:    
            return filename  
    except:  
        print "ERROR"  
        sys.exit()

Hello all…i have a problem with the code above. When i call the ‘checkarg’ function, if i did not pass any parameter on the command line i have the “ERROR” output and sys exit, just as expected.

But, if i provide a parameter on the command line (like “-mycommand”) it prints the “SPECIFIC_TEXT” and then prints “ERROR” message from the EXCEPT block too.

The TRY block will only run when I provide a parameter, if I don’t, then EXCEPT will get the turn. But, it is running the TRY and EXCEPT blocks together.

Does anybody knows the reason of this behavior?? Any mistake on my code? Tks for all !

Answer 1

I think I understand your question…

sys.exit() exits by raising a SystemExit exception, which your except statement is catching.