• Lutger (22/22) Aug 21 2006 I have a basic, perhaps even dumb question that came up with the awesome...
• Chris Nicholson-Sauls (31/57) Aug 21 2006 I tried it myself with DMD 0.165 as:
• Lutger (3/69) Aug 21 2006 Great, thank you. I agree that this line you changed looks much better:

Lutger <lutger.blijdestijn gmail.com> writes:

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.

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;
}

Chris Nicholson-Sauls <ibisbasenji gmail.com> writes:

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:





















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

Lutger <lutger.blijdestijn gmail.com> writes:

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:




















 
 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