On 14/06/18 08:04, Deepak Dixit wrote: > def test2(nums=[]): > nums.append(len(nums)); > return nums > > print 'test2()', test2() > print 'test2([1,2,3])', test2([1,2,3]) > print 'test2([1,2])', test2([1,2]) > print 'test2()', test2() > print 'test2()', test2() > > Calling test2 > ======================================== > test2() [0] > test2([1,2,3]) [1, 2, 3, 3] > test2([1,2]) [1, 2, 2] > test2() [0, 1] > test2() [0, 1, 2] > > I am assuming that in test1() we are not doing any modification in the > passed list and because of that its working as per my expectation.
Correct. You are still dealing with 2 separate objects(see below) but you don't modify anything so it looks like it works as you expected. > But when I am calling test2() with params and without params then both are > using different references. Why ? Can you please help me to understand > this. When you create a default value for a function parameter Python creates an actual object and uses that object for every invocation of the function that uses the default. If the object is mutable (a list or class instance, say) then any modifications to that object will stick and be present on subsequent calls. Thus in your case the first call of the function the list is empty and has len() 0, the second call it still has the zero stored so len() is now 1, and so on. But when you call the function without using the default Python does not involve the default object at all, it uses whichever object you provide. The same applies to your tst1 function. Each time you use the default call the same empty list object is being printed. But because you don't modify it it always appears as an empty list. -- Alan G Author of the Learn to Program web site http://www.alan-g.me.uk/ http://www.amazon.com/author/alan_gauld Follow my photo-blog on Flickr at: http://www.flickr.com/photos/alangauldphotos _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: https://mail.python.org/mailman/listinfo/tutor
