let (x,y) = ...

[visitor via Digitalmars-d-announce]

On Monday, 23 November 2015 at 10:28:53 UTC, thedeemon wrote:
> On Sunday, 22 November 2015 at 18:47:34 UTC, visitor wrote:
>
>> What is the reason for using pointers (alias pointerOf(T) = T*
>>  etc...)
>> it works without ! what am i missing ?
>
> What and how exactly works without?
> My original solution remembers in the constructor addresses of 
> variables to fill, then does the filling in opAssign operator, 
> so I needed a way to store the references and used pointers for 
> that.

yes, but you are using ref : "auto let(Ts...)(ref Ts vars)"
so vars are changed, no need to store anything, no?
i was wondering if there is some subtleties or efficiency reasons 
for using pointers

this work fine with your unittest :
auto let(Ts...)(ref Ts vars) {
     struct Let
     {
         void opAssign( Tuple!Ts xs ) {
             foreach(i, t; Ts)
                 vars[i] = xs[i];
         }

         static if (sameTypes!Ts) {
             import std.conv : text;
             void opAssign(Ts[0][] xs) { // redundant but more 
effective
                 enforce(xs.length == Ts.length, "let (...) = ...: 
array must have " ~ Ts.length.text ~ " elements.");
                 foreach(i, t; Ts)
                     vars[i] = xs[i];
             }

             void opAssign(R)(R xs) if (isInputRange!R && 
is(ElementType!R == Ts[0])) {
                 static if (hasLength!R) {					
                     enforce(xs.length >= Ts.length, "let (...) = 
...: range must have at least " ~ Ts.length.text ~ " elements.");
                 }
                 foreach(i, t; Ts) {
                     enforce(!xs.empty, "let (...) = ...: range 
must have at least " ~ Ts.length.text ~ " elements.");
                     vars[i] = xs.front;
                     xs.popFront();
                 }
             }

             void opIndexAssign(R)(R xs) if (isInputRange!R && 
is(ElementType!R == Ts[0])) {
                 foreach(i, t; Ts) {
                     if(xs.empty) return;
                     vars[i] = xs.front;
                     xs.popFront();
                 }
             }
         }
     }
     return Let();
}