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Right now I think I have understood most of the problems related to default value in Python. The questions on Stackoverflow has been a great help, but there is one case/example that they do not address. That example is in this link: http://effbot.org/zone/default-values.htm Here is the part in question:

Finally, it should be noted that more advanced Python code often uses this mechanism to its advantage; for example, if you create a bunch of UI buttons in a loop, you might try something like:

for i in range(10): def callback(): print "clicked button", i UI.Button("button %s" % i, callback)

only to find that all callbacks print the same value (most likely 9, in this case). The reason for this is that Python’s nested scopes bind to variables, not object values, so all callback instances will see the current (=last) value of the “i” variable. To fix this, use explicit binding: 

for i in range(10): def callback(i=i): print "clicked button", i UI.Button("button %s" % i, callback)

The i=i part binds the parameter i (a local variable) to the current value of the outer variable i. I can understand why the second example works the way it does: in the def line, the i in the function is explicitly bounded to the i in the for (this is done during the "def" statement, namely definition-time). But what I don't get is how the first example works. It seems to me that the i variable there is bounded during run-time but then shouldn't it also be the same as the second one? Thank you very much.

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  • @TigerhawkT3 : No, this is not a duplicate Your link only mention an example of default value without a loop, while my question is concerned with behaviour of a function in a loop. These are 2 different examples. Commented Oct 18, 2015 at 5:06
  • Then try stackoverflow.com/questions/19837486/python-lambda-in-a-loop. Commented Oct 18, 2015 at 5:11
  • Read this answer in particular. Commented Oct 18, 2015 at 5:12
  • I think Tigerhawk's link is probably the most descriptive. Essentially you've defined a function that reads i, but only does so at runtime. What i is at time of definition is unimportant. If you move the assignment i=i into the def statement, then Python will intern that value as the local i (since the whole def line gets executed at definition time, not at run time) Commented Oct 18, 2015 at 5:22
  • @AdamSmith: The link's examples ares somewhat different from mine. In the first example, f() is called after the for loop, which makes sense because funcs.append(lambda: x) would then call the value x which is at this time, 3. But in my example, i is called during the for loop, not after it, therefore it doens't make sense for i to be 9 all the time. This is what confuses me. I think this question deserves to be answered in it's own right. Commented Oct 18, 2015 at 8:24
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