I am sorry --- I did not realize you had anger management issues. In
any case, consider the example in the section of the manual I linked,
with macros and functions:
macro time_mac(ex)
return quote
local t0 = time()
local val = $ex
local t1 = time()
println("elapsed time: ", t1-t0, " seconds")
val
end
end
function time_fun(ex)
return quote
local t0 = time()
local val = $ex
local t1 = time()
println("elapsed time: ", t1-t0, " seconds")
val
end
end
Then observe the output of
time_fun(:(foo()))
macroexpand(:(@time_mac(foo()))
On Sun, Jun 05 2016, Ford O. wrote:
> Don't make me angry.
>
> I have read that chapter more than 5 times.
> You could also read my question 5 times.
>
> I am asking why do we have to specify macro and function with different
> keyword, when compiler exactly knows whether we are calling macro or
> function.
>
> Example for ...:
> callable foo(e::Expr) = :(&e)
> @foo 2 + 2
> foo(2 + 2)
> # two versions of foo callable are compiled, one as macro, one as function
>
> Does the *macro *keyword have any significance for compiler?
>
> On Sunday, June 5, 2016 at 1:38:32 PM UTC+2, Tamas Papp wrote:
>>
>> Sorry for linking a local version of the manual -- still, you could
>> probably have found it with a bit of effort, but it appears that you are
>> adamant in your refusal to read it.
>>
>> I would say that the most important reasons is that macros have special
>> conventions because of hygiene, see
>> http://docs.julialang.org/en/release-0.4/manual/metaprogramming/#hygiene
>>
>> Some languages (eg Common Lisp) have the same namespace for macros and
>> functions, Julia distinguishes them with @. So they are not as seamless
>> as in CL, but at the same time they stand out and warn the reader that
>> source transformation is going on. There could be arguments for both
>> choices, but Julia chose this one.
>>
>> On Sun, Jun 05 2016, Ford O. wrote:
>>
>> > You got it wrong.
>> > In different words.
>> >
>> > Why do we need to specify macro and function block with macro and
>> function
>> > keyword? Isnt the '@' symbol enough?
>> >
>> > On Sunday, June 5, 2016 at 1:11:48 PM UTC+2, Tamas Papp wrote:
>> >>
>> >> A macro is a function that is used to transform (a representation) of
>> >> source code. Consequently, it is called when Julia evaluates your
>> >> function defintions, not at runtime. Please read
>> >> file:///usr/share/doc/julia-doc/html/manual/metaprogramming.html where
>> >> you find examples.
>> >>
>> >> Regarding you example: it is unclear what you are trying to do, but
>> >> unless you use the macro keyword, you get an ordinary function, not a
>> >> macro.
>> >>
>> >> On Sun, Jun 05 2016, Ford O. wrote:
>> >>
>> >> > What makes macro different from function?
>> >> >
>> >> > Why is it not possible to do:
>> >> > foo(e::Expr) = :(&e)
>> >> > @foo x = x + 5
>> >> >
>> >> > On Friday, June 3, 2016 at 10:05:46 AM UTC+2, Ford O. wrote:
>> >> >
>> >> > I think this deserves an own topic.
>> >> >
>> >> > You should post here syntax that looks like duplicate or you think
>> it
>> >> could use already existing keyword. (mark with # identical or #
>> >> replacement)
>> >> > Rule of thumb - the less special syntax the better.
>> >> >
>> >> > # identical
>> >> > # replace ' , ' with ' ; ' in arrays ?
>> >> > [1, 2, 3, 4]
>> >> > [1; 2; 3; 4]
>> >> >
>> >> > # identical
>> >> > # replace ' = ' with ' in ' in for loops ?
>> >> > for i = 1:10
>> >> > for i in 1:10
>> >> >
>> >> > # replacement
>> >> > # replace ' -> ' with ' = ' in anonymous functions ?
>> >> > (a, b, c) -> a + b + c
>> >> > (a, b, c) = a + b + c
>> >>
>> >>
>>
>>
