[EMAIL PROTECTED] wrote:
> ####################
> Now, CPS would transform the baz function above into:
>
> def baz(x,y,c):
> mul(2,x,lambda v,y=y,c=c: add(v,y,c))
>
> ###################
>
> What does "y=y" and "c=c" mean in the lambda function?
they bind the argument "y" to the *object* currently referred to by the
outer "y" variable. for example,
y = 10
f = lambda y=y: return y
y = 11
calling f() will return 10 no matter what the outer "y" is set to.
in contrast, if you do
y = 10
f = lambda: y
y = 11
calling f() will return whatever "y" is set to at the time of the call.
or in other words, default arguments bind to values, free variables bind
to names.
> I thought it bounds the outer variables, so I experimented a little
> bit:
>
> #################
> x = 3
> y = lambda x=x : x+10
>
> print y(2)
> ##################
>
> It prints 12, so it doesn't bind the variable in the outer scope.
it does, but you're overriding the bound value by passing in a value. try:
x = 3
y = lambda x=x : x+10
y()
x = 10
y()
instead.
</F>
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