Thanks for your reply. I guess I was forgetting that at macroexpansion time
(if I'm using the word correctly), we don't have access to the binding
between u and 10. The same error I message I noted above does appear for (+
2 'u).
And, if I understand correctly, what was really happening with the macro
bar was that in ('+ 2 'u), '+ was looking itself up in 2, not finding
itself, and so returning the default value 'u. This was the expansion of
the macro, and so at run time, it was bound to 10.
I think.
On Tuesday, January 26, 2016 at 2:02:22 PM UTC-8, Dan Burton wrote:
>
> > I know that this macro will add 2 to its argument:
>
> This isn't what that macro does. That macro takes a piece of syntax, and
> sticks that syntax in the hole in the expression (+ 2 HOLE). The
> macro doesn't evaluate the expression, it just generates it. It is the
> evaluation of *that* expression that yields the plus-two result.
>
> The arguments to a macro are not values, they are syntax. The result of a
> macro isn't a value, it's syntax. When you call (foo u), you are not
> passing the value of u into the macro. You are passing in the syntactic
> symbol u. That's why it says you can't add a Symbol to a Number. Your macro
> is trying to compute (at "compile time") (+ 2 'u).
>
> On Tuesday, January 26, 2016, <echigoy...@gmail.com <javascript:>> wrote:
>
>> I'm new to clojure and macros and am having trouble wrapping my head
>> around some simple things.
>>
>> I know that this macro will add 2 to its argument:
>>
>> user=> (defmacro plus-two [x] `(+ 2 ~x))
>> #'user/plus-two
>> user=> (plus-two 5)
>> 7
>> user=> (def u 10)
>> #'user/u
>> user=> (plus-two u)
>> 12
>>
>> I tried the following just to see what will happen and don't understand
>> what's going on:
>>
>> user=> (defmacro foo [x] (+ 2 x))
>> #'user/foo
>> user=> (foo 5)
>> 7
>> user=> (foo u)
>>
>> ClassCastException clojure.lang.Symbol cannot be cast to
>> java.lang.Number clojure.lang.Numbers.add (Numbers.java:128)
>>
>> I tried quoting the plus:
>>
>> user=> (defmacro bar [x] ('+ 2 x))
>> #'user/bar
>> user=> (bar 5)
>> 5
>> user=> (bar u)
>> 10
>>
>> This makes sense, since (bar u) is the same as + 2 u (without
>> parentheses), which returns u. But I don't understand what's happening with
>> foo, and I can't check it with macroexpand since that gives the same error.
>> Is there a way to recreate this error without defining a macro?
>>
>> Thanks in advance for any and all responses.
>>
>>
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>
>
> --
> -- Dan Burton
>
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