> | > | Starting point: > | ... > | self.status['text'] = 'Processing ...' > | try: > | cli_main(argv) > | except Exception, e: > | self.status['text'] = 'Error: ' + str(e) > | return > | ... > | cli_main: > | > | keypath, inpath, outpath = argv[1:] > | ... > | with open(inpath, 'rb') as inf: > | serializer = PDFSerializer(inf, keypath) > | with open(outpath, 'wb') as outf: > | filenr = outf.fileno() > | serializer.dump(outf) > | return 0 > | > | PDFSerializer.dump: > | > | def dump(self, outf): > | self.outf = outf > | ... > > See that you set serializer.outf to the outf you open in cli_main? > Any attempt to use serializer _after_ exiting the "with open(outpath, > 'wb') as outf" will use serializer.outf, but the outf is now closed. > And thus itsfiledescriptoris invalid.
Okay, I changed it to a try: ... finally: block where I open the file and in finally I close it. Nothing has changed. The error still occures. Doesn't the with open(outpath, 'wb') as outf: clause has to wait until the pdfserialiser.dump method has finished anyway? IMHO it can't just call it and immediately close it. At least the try: finally: construct should work? Or does it the same (call the method and immediately jump to the finally close)? Would it work if I would write: with closing(outpath, 'wb') as outf: ? I'm a little bit confused about Python's strange processing ... -- http://mail.python.org/mailman/listinfo/python-list
