On 01/-10/-28163 02:59 PM,
mhearne808[insert-at-sign-here]gmail[insert-dot-here]com wrote:
I am just trying to wrap my head around decorators in Python, and I'm
confused about some behavior I'm seeing. Run the code below (slightly
adapted from a Bruce Eckel article), and I get the following output:
inside myDecorator.__init__()
inside aFunction()
Finished decorating aFunction()
inside myDecorator.__call__()
My question: Why isn't the first print statement in "__main__" the
first line of code executed? Is aFunction() not closed somehow?
#!/usr/bin/env python
class myDecorator(object):
def __init__(self, f):
print "inside myDecorator.__init__()"
f() # Prove that function definition has completed
def __call__(self):
print "inside myDecorator.__call__()"
@myDecorator
def aFunction():
print "inside aFunction()"
if __name__ == '__main__':
print "Finished decorating aFunction()"
aFunction()
classes and functions and decorators have some portions that execute
when they occur, long before anybody "calls" them. (I'm sure there are
other examples; one might consider imports the same way)
In the case of classes, anything outside of the method definitions will
happen before the class definition is completed. For example, class
attributes happen at that time.
For functions/methods, default arguments are evaluated at the definition
time. So if the default value makes a call, the call will happen at
that time.
Function decorators execute right after the corresponding function
definition is built. Such decorators won't normally call the function,
but as you notice, if you do call it, it will execute.
When you think about it, these behaviors are the only reasonable way
these things could be done, unless the compiler tried to do some "just
in time" compiling, not really building the code till somebody uses it.
And that would make the language a lot different.
DaveA
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