How to expand an expression along with a parameter tuple?

[Ali Çehreli]

On 06/17/2013 02:32 AM, TommiT wrote:

 > On Monday, 17 June 2013 at 07:20:23 UTC, Ali Çehreli wrote:
 >>
 >> The following does not answer the question of expanding but at least
 >> foo() receives [30, 70, 110] :)
 >>
 >> import std.stdio;
 >> import std.algorithm;
 >> import std.array;
 >> import std.range;
 >>
 >> int[] arr = [ 1, 3, 5, 7, 11 ];
 >>
 >> void foo(T)(T[] values...)
 >> {
 >>     writeln(values);
 >> }
 >>
 >> void bar(T)(T[] values...)
 >> {
 >>     foo(arr
 >>         .indexed(values)
 >>         .map!(a => a * 10)
 >>         .array);
 >> }
 >>
 >> void main()
 >> {
 >>     bar(1, 3, 4);
 >> }
 >>
 >> Ali
 >
 > Yeah, that would work. I'd hate the overhead though.

There is no inherent overhead though. I called .array only because I 
thought that the C++ version was eager. If we stay lazy and D-like, we 
can make foo() take a range:

import std.stdio;
import std.algorithm;
import std.range;

int[] arr = [ 1, 3, 5, 7, 11 ];

void foo(R)(R range)          // <-- now takes range
{
     writeln(range);
}

void bar(T)(T[] values...)
{
     foo(arr
         .indexed(values)
         .map!(a => a * 10));  // <-- no .array anymore
}

void main()
{
     bar(1, 3, 4);
}

I hear that ldc (and perhaps gdc) perform pretty magical compiler 
optimizations. There is no reason for the code above to be slower than 
the hand-written equivalent.

Ali