Hello! I’m newcomer to Python and I’m on documentation reading stage and trying some of examples. I’m using Win7 x64 OS and Python 2.7.3 (default, Apr 10 2012, 23:24:47) [MSC v.1500 64 bit (AMD64)]. I try to understand how string format expression (%)works. Everything is almost clear but except one case: using ofg(G) conversion type and # flag. Let’s take a look at documentation here: http://docs.python.org/release/2.7.3/library/stdtypes.html#sequence-types-str-unicode-list-tuple-bytearray-buffer-xrange Document declares for g(G) conversion type in case of using # flag (4th note): “The precision determines the number of significant digits before and after the decimal point and defaults to 6”.
I have noticed behavior that does not meet documentation declaration and looks like a bug in case when using g(G) conversion type with # flag with omitted precision and zero integer part of the decimal. Could someone, please comment the case it is a bug or right use case result? If it is correct, please explain why. Steps to reproduce the case: 1.Start python interactive mode 2.Enter string with g(G) conversion type and using #flag like this: "%#g"%0.3 – precision parameter is omitted and integer part of the decimal is zero. 3.Watch the output results Actual result: Python outputs decimal as declared as but with more significant digits than default value of 6 - if integer part of the decimal is equal to zero. >>> "%#g"%0.3 '0.300000' >>> "%#G"%0.3 '0.300000' >>> "%#G"%0.004 '0.00400000' >>> Expected results: As declared in documentation – there will be 6 significant digits before and after decimal point by default. Thanks, Regards, Lesya. _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
