best way to increment an IntVar?

[Dave Angel]

Alan G Isaac wrote:
> On 6/25/2010 3:52 PM, Dave Angel wrote:
>> I said "default", not "only" behavior.   I suspect list provides an
>> __iadd__  method to provide this ability.  Integers do not, and
>> therefore neither does the object the OP was asking about.
>
>
> I have no idea what "default behavior" is supposed to mean.
> Mutable objects like a list will generally modify in place.
> Immutable objects of course will not.  An IntVar is mutable.
> You have given no reason for it not to handle ``+=`` in an
> unsurprising fashion.  It is not an int.
>
> Alan Isaac
>
>
To quote the docs,

"If a specific method is not defined, the augmented assignment falls 
back to the normal methods. For instance, to execute the statement x += 
y, where /x/ is an instance of a class that has an __iadd__() 
<#object.__iadd__> method, x.__iadd__(y) is called. If /x/ is an 
instance of a class that does not define a __iadd__() <#object.__iadd__> 
method, x.__add__(y) and y.__radd__(x) are considered, as with the 
evaluation of x + y."

In other words, if a class defines __add(), but not __iadd__(), then 
it'll behave this way.  That's what I mean by default.

I didn't say it should, merely that it does.  I suspect that __iadd__ 
didn't exist when IntVar was being defined, and nobody got around to 
adding it.

Another part of the docs would seem to encourage IntVar to change:

"An augmented assignment expression like x += 1 can be rewritten as x = 
x + 1 to achieve a similar, but not exactly equal effect. In the 
augmented version, x is only evaluated once. Also, when possible, the 
actual operation is performed /in-place/, meaning that rather than 
creating a new object and assigning that to the target, the old object 
is modified instead"

DaveA