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(I can work around this but I am curious as to why it happens.)

I have the standard installation of Python on Windows 10. I also have the Anaconda package installed.

I noticed that when I started IDLE from the start menu it launched thus

Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 07:18:10) [MSC v.1900 32 bit (Intel)] on win32

Yet when I go to the actual folder the link points to, and open idle.pyw it starts thus

Python 3.6.5 |Anaconda, Inc.| (default, Mar 29 2018, 13:32:41) [MSC v.1900 64 bit (AMD64)] on win32

Strange...

What exactly is causing it to start in these different ways?

The only way I can open the Python 3.6.0 is now through the start menu link - how could I launch 3.6.0 without having to go through the start menu?

iansedano
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  • After having this happen to me again, while using CYGWIN, in which case, it was a real problem because suddenly cygwin was unable to do the most basic things like cd! I suppose it is to do with where the program is called from. If it is called from a .lnk file from the start menu, then it seems windows automatically chooses the right folder to launch it from, as if in command line. Yet if you go into the file, then it seems like it calls the file from the wrong place and so the resources are assigned in a strange way. – iansedano Nov 08 '18 at 09:56

1 Answers1

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Either you have misinterpreted something or there is something strange in the registry. In either case, once you have the 32-bit 3.6.0 open, you can right click on the icon in the taskbar and click 'Pin to taskbar'. Or, you can drag and drop the Start Menu icon to the desktop. Either way, I recommend upgrading to 3.6.6 now or 3.6.7 in a week or so.

Terry Jan Reedy
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