Is it possible to quote code where the methods are already resolved? For
example, instead of `+` being a symbol in the quoted code, it would be
resolved to the + operator from the current module?
On Friday, May 23, 2014 8:31:19 PM UTC-4, Isaiah wrote:
>
> parse("...codez...") can be helpful too.
>
> For (1) & (2), the single-quoted numbers are interpreted as literals
> rather than symbols. It is possible to create a Symbol called `:1` with
> `symbol("1")`, if needed, but that seems strange. The entry syntax `:1`
> might be disallowed for efficiency and disambiguation - I'm not sure.
>
> For (3), in general, a TopNode tells the compiler to resolve any names to
> the current base module. It might be illustrative to look at the difference
> between these expressions:
>
> julia> a = 1
> 1
>
> julia> :($a + 1)
> :(1 + 1)
>
> julia> :(1+:(a+2))
> :(1 + :(a + 2))
>
> julia> :(1+:($a+2))
> :(1 + top(Expr)(:call,:+,a,2))
>
> The last case asks for an interpolation which cannot be expanded at parse
> time, because the interpolation is quoted. The `top(Expr)` and `call` are
> generated to guarantee that `foo` - whatever the value of `foo` ends up as
> in the Top module namespace - is used when the expression is evaluated.
>
> (as an aside, eval'ing those last two examples will give errors,
> because it is not possible to add an Int and an Expr)
>
>
>
>
>
> On Fri, May 23, 2014 at 3:53 PM, Adam Smith <swiss.arm...@gmail.com
> <javascript:>> wrote:
>
>> I find it quite helpful to use dump() on expressions to see the first few
>> levels of the parsed AST. That makes it easier to see what Julia is doing
>> with your expressions.
>>
>>
>> On Thursday, May 22, 2014 5:17:13 PM UTC-4, Andrew McKinlay wrote:
>>>
>>> I have been trying to grok Julia macros, so I've dived into the
>>> metaprogramming docs. Whilst playing around with the interpreter, I've run
>>> into some peculiarities that I don't quite understand.
>>>
>>> julia> versioninfo()
>>> Julia Version 0.3.0-prerelease+2690
>>> Commit e4c2f68* (2014-04-20 12:15 UTC)
>>> ...
>>>
>>>
>>> 1. Why is typeof( :(:(1+2)) ) == Expr, but typeof( :(:(1)) ) ==
>>> QuoteNode?
>>> 2. Why is typeof( :(1) ) == Int64, but again typeof( :(:(1)) ) ==
>>> QuoteNode?
>>> 3. Why is typeof( :(1+:($1+2)).args[3].args[1] ) == TopNode?
>>>
>>>
>>> I was assuming that quoting was just a shorthand way for constructing
>>> expressions, since the docs say:
>>>
>>>> There is special syntax for “quoting” code (analogous to quoting
>>>> strings) that makes it easy to create expression objects without
>>>> explicitly
>>>> constructing Expr objects.
>>>
>>> But the types of objects constructed by quoting do not always take the
>>> form of an Expr, as above.
>>>
>>> What is going on here?
>>>
>>
>
