Stuart Sierra wrote:
> > - The reader performs a few substitutions, then gives the resulting
> > code to the compiler.
> > - The compiler starts by expanding macros into their substitution.
> > --- Thus, for functions, there's no such thing as macros.
> > --- Thus, all that functions see of macros is what came out of macros'
> > complete expansion.
> > --- Therefore, like Timothy Pratley wrote so well, a function can have
> > for parameter a macro call, but not a macro itself.
>
> All correct up to this point.
>
> > - After macro expansion, the compiler starts compiling all functions
> > in need of being compiled, including the functions modified during
> > macro expansion.
>
> Sort of. What's really getting compiled are just the function calls.
> The functions themselves were compiled back when they were defined.
Ok, so do we have: read => macro-expand => compile-definitions =>
compile-calls? So that when new code is read, if it contains function
definitions, they'll get compiled before their call(s) is compiled?
Stuart and Konrad respectively said:
> [...] "and" doesn't know how many arguments it's going to get.
> If you look at the definition of "and", you'll see that it takes a
> variable number of arguments and it is recursive. So if you tried to
> expand it before knowing the length of the argument list, you would
> get an infinite loop.
> > > The universal workaround is to define a
> > > function that contains nothing but the macro, such as
> > > (fn [x y] (and x y))
> > > which can also be written in Clojure using the shorthand notation
> > > #(and %1 %2)
Ok!! So *that* would explain why wrapping 'and' *this* way enables it
to be an argument to a function call in (reduce #(and % %2): it
becomes clear how many args are going to be given to 'and's call. Wow,
thanks for such clarity!
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