On Tuesday, 19 July 2016 at 17:05:55 UTC, Rufus Smith wrote:
On Tuesday, 19 July 2016 at 16:59:48 UTC, Lodovico Giaretta
wrote:
On Tuesday, 19 July 2016 at 16:50:56 UTC, Rufus Smith wrote:
On Tuesday, 19 July 2016 at 16:09:38 UTC, Lodovico Giaretta
wrote:
[...]
But this doesn't create a function with all the attributes of
the original? Just one that has the same return type and
parameters. What if Fun is pure or extern(C) or some other
attributes? I'd like to create a function that is exactly the
same in all regards as the original.
Sorry, I misunderstood your question.
With the method I showed you, if the function is @safe, pure,
@nogc or nothrow, foo will infer those attributes. But only if
the operations you do in foo (apart from calling bar) are
themselves @safe, pure, @nogc or nothrow.
For other things, like extern(C), I don't think there's a
simple solution; but I'm not an expert, so I hope someone else
will give you a better answer.
What is strange is I cannot even pass an extern(C) function to
foo.
void foo(R, A...)(R function(A) bar);
extern(C) void bar();
foo(&bar)
fails. Remove extern and it passes. I have not figured out how
to allow for extern(C) functions to be passed.
That's because an extern function must be called with a different
code. So it cannot be cast to a non-extern(C) function pointer,
which is what your foo accepts. If you follow my advice, and make
the entire function type a parameter of foo, then foo will at
least accept your extern(C) function, but it will not be
extern(C) itself.
