I don't think you can do that ([] not ()). But what I showed you would work well enough.
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What syntax for 'glob' argument?
I want to load some specific files in a directory. The filter on the filename would be ideally a regular expression. However it seems glob argument is not a regular expression... while not being a standard string as well since their is still some level of interpretation.
E.g, I can load both C headers and files using glob('*.[ch]'). But how to load both *.h and *.cpp files -- other than calling glob twice. What are the interpretation rules of glob argument, that make '[]' interpreted but not e.g. '()'? I'd appreciate any help or pointers.
Thanks,
Pascie
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Answers
Okay, I have to use '{}' and not '()'. I guess that's because that is the csh standard, according to perldoc -f glob.
glob("*.h *.cpp") seems to have a stange behavior, at least on windows.
glob('c:\\mydir\\*.h') would give the *.h in mydir, but
glob('c:\\mydir\\*.h c:\\mydir\\*.cpp') would give the *.h and *.cpp in the working directory...
(using perl 5.8.0 of activestate under win2000)
Thanks!
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accept answer by thoellri.
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by: inq123Posted on 2003-09-12 at 12:25:49ID: 9348319
glob("*.h *.cpp");