Why doesn't __call__ lead to infinite recursion?

John J. Lee jjl at pobox.com
Sat Aug 16 13:19:50 CEST 2003

aahz at pythoncraft.com (Aahz) writes:
> you'll see __call__ there.  When you do foo(), Python actually does
> type(foo).__call__(foo).  Because type(foo).__call__ is manipulating
> foo, you don't get the circular reference.

Still seems weird that type(foo).__call__.__call__ (etc.) is defined.


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