Fred Yankowski writes:

.... <http://www.digicool.com/releases/ExtensionClass/Acquisition.html>? I think I understand how the various "complex expressions" relate to the equivalent expression in terms of the '__of__' operator, except for the very last case, "a.b.c.x". I just can't follow why the equivalent expression isn't

x.__of__(a).__of__(c.__of__(b.__of__(a)))

rather than the more complex answer given:

x.__of__(a).__of__(b.__of__(a)).__of__(c.__of__(b.__of__(a))) Acquisition wrappers are ExtensionClass instances with an "__of__" method.

Whenever, you look up an attribute in an ExtensionClass instance, "i", and the directly looked up object, "o", (!) has an "__of__" method, the result is not "o" but "o.__of__(i)". In your example, "x" is looked up in "c.__of__(b.__of__(a))". If we assume, that "x" has a "__of__" method, the result will be: "x1.__of__(c.__of__(b.__of__(a))" with "x1" the result of directly looking up "x" in "c.__of__(b.__of__(a))". "x" is first looked up in "c". As it is not found there, it is looked up in "b.__of__(a)". Thus, we have "x1=(b.__of__(a)).x" As "b.__of__(a)" is an ExtensionClass instance (and we assumed "x" has an "__of__" method). The result will be "x2.__of__(b.__of__(a))" with "x2" the result of directly looking up "x" in "b.__of__(a)". "x" is first looked up in "b". As it is not found there, it is looked up in "a". Thus, we have "x2= a.x" If "a" is an ExtensionClass instance, then "a.x" is "x.__of__(a)". Put things together and you will get the complex result. Dieter