Is this an example of tail recursion?
rosuav at gmail.com
Wed Aug 5 18:28:57 CEST 2015
On Thu, Aug 6, 2015 at 2:10 AM, Rustom Mody <rustompmody at gmail.com> wrote:
> 1 + x
> does not *call* 1 .__add__(x)
> It *is* that
> [Barring corner cases of radd etc]
> IOW I am desugaring the syntax into explicit method-calls so you can see
> all the calls explicitly
> Then it becomes evident -- visibly and in fact --that the tail call is the
> __add__ method not the solderdiersVsDefenders
Except that it *isn't* that, precisely because of those other cases.
When Python sees an expression like "1 + x" and doesn't yet know what
x is, it can't do anything other than record the fact that there'll be
a BINARY_ADD of the integer 1 and whatever that thing is. That object
might well define __radd__, so the call is most definitely not
equivalent to the operator.
And the ultimate result of that addition might not even be a function
call at all, if it's implemented in C. Or if you're running in PyPy
and the optimizer turned it into machine code. So no, even though you
can define addition for *your own classes* using __add__ or __radd__,
you can't reinterpret every addition as a function call.
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