>
> Macroexpansion is part of the expression evaluation mechanism. In your
> example
>
>
> (defn foo
> ([bar] (foo bar :default))
> (special-fn-spec))
>
> the whole defn-form is macroexpanded first, yielding something like
>
> (def foo (fn ([bar] (foo bar :default)) (special-fn-spec))
>
> Then the fn form is evaluated, yielding a compiled function. At that point,
> the compiler checks its syntax, and finds two bodies, one well-formed (arg
> list followed by expression) and a second ill-formed one (just an
> expression). The function bodies are *not* macroexpanded because they are
> not evaluated either. The only other subform of your example that is ever
> macroexpanded is (foo bar :default).
>
> There are a couple of ways to generate function bodies programmatically,
> but it is difficult to give useful advice without knowing what you need this
> for.
>
> Konrad.
Thanks Konrad, that makes sense.
I suppose I was a bit confused about when macroexpansion occurs.
My real use-case involves wrapping a Java object, which has a number of
methods with varying numbers of optionally nullable parameters. E.g.
DatabaseMetaData#getExportedKeys(String, String, String)
In this method, the first two parameters may be null. So, my fn looks like
this:
(defn exported-keys
([table]
(exported-keys nil table))
([schema table]
(exported-keys nil schema table))
([catalog schema table]
(fetch-metadata-rs .getExportedKeys catalog schema table)))
I was trying to automatically generate the final function body since it
duplicates the parameter list, and I expect to have a lot of these kinds of
methods. Although, it's not too bad as it is (I've already pulled some
common bits into fetch-metadata-rs).
Not sure if that makes sense or not... ?
Cheers,
Stuart
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