Evaluation of expressions and the comma operator

[Lutger]

Great, thank you. I agree that this line you changed looks much better: 
T[] result = new T[count];

Chris Nicholson-Sauls wrote:
> Lutger wrote:
>> I have a basic, perhaps even dumb question that came up with the 
>> awesome new implicit conversion of expressions to delegates:
>>
>> In comma seperated expressions, (how) is the result of evaluation 
>> defined? For example does (a, b) always yields the value of b? I 
>> expect so, but I want to be sure.
> 
> Yes.  The value of a Comma is always the result of the last expression 
> in the sequence.
> 
>> Maybe some code is clearer, I was toying around to do this:
>>
>> void main()
>> {
>>     int a = 0;
>>     int b = 1;
>>     int c = 0;
>>
>>     // Prints a fibbonaci sequence, is this legal?
>>     writefln( generate(10, ( c = (a + b), a = b, b = c) ) );
>> }
>>
>> T[] generate(T)(int count, T delegate() dg)
>> {
>>     T[] array;
>>     array.length = count;
>>     foreach(inout val; array)
>>         val = dg();
>>     return array;
>> }
> 
> I tried it myself with DMD 0.165 as:
> # import std .stdio ;
> #
> # void main () {
> #   int a = 0 ,
> #       b = 1 ,
> #       c = 0 ;
> #
> #   // prints Fibonacci sequence
> #   // Note: this DOES NOT include the natural [0,1] prefix subsequence
> #   writefln(generate(10, (c = (a + b), a = b, b = c)));
> # }
> #
> # T[] generate (T) (int count, T delegate() expr) {
> #   T[] result = new T[count];
> #
> #   foreach (inout val; result) {
> #     val = expr();
> #   }
> #   return result;
> # }
> 
> The result was a perfect compilation, and when ran it printed:
> [1,2,3,5,8,13,21,34,55,89]
> 
> So it works without a hitch, except for the failure to include the [0,1] 
> at the beginning of the Fib sequence... but that's not such a huge 
> deal.  You can just define a
>     const int[] FIB_PRE = [0, 1];
> 
> And call it as
>     writefln(FIB_PRE ~ generate(10, (c = (a + b), a = b, b = c)));
> 
> -- Chris Nicholson-Sauls