On Tuesday, 1 September 2015 at 21:17:10 UTC, jmh530 wrote:
Consider these three different ways to import std.stdio
import std.stdio;
import std.stdio : writeln;
static import std.stdio;
and suppose writeln is the only function in std.stdio the
program is using.
In each case, the size of the exe is exactly the same. So
presumably, the compiler is smart enough to see that
std.stdio.chunks, for instance, isn't used and not to include
in the final program. However, there might be a difference in
compile times (though I don't exactly know how to measure that).
My understanding is that import std.stdio; leads to all of
std.stdio being compiled, but import std.stdio : writeln leads
to only writeln being compiled (and anything it depends on).
There are also faster lookups during compilation, or something,
I think. Given that the documentation only indicates the
difference between import std.stdio and static import std.stdio
is the ambiguous-ness that you point out, I would expect that
the time to compile would be the same as import std.stdio. Is
this correct?
Oh no. Importing in D is a symbol table import. If you're using a
module and it's only available in source-format to the compiler,
then it has to compile the whole module even if you're using just
one function (because internally the whole module is likely to be
very tightly coupled, and usually should be, if it's under one
name).
import std.stdio; // makes available for calling all symbols
in std.stdio; compiles module if not pre-compiled and if at least
one symbol is used (not sure about the last point)
import std.stdio : writeln; // makes available for calling
only 'writeln()' from std.stdio; compiles module if not
pre-compiled
static import std.stdio; // makes available for calling all
symbols in std.stdio but only when fully qualified; compiles
module if not pre-compiled
As I understand it, the only difference between cases with
different numbers of called symbols from a module is apparent
during linking, not compilation.
Anyone can shed light on this?
