• Temtaime (8/8) Jun 09 2013 @(3) int a;
• David (9/19) Jun 09 2013 It works like that:
• Temtaime (2/2) Jun 09 2013 Oh, thanks. But then that code doesn't work:
• Temtaime (1/1) Jun 09 2013 Thanks for all, it's unactual anymore.
• Jacob Carlborg (7/14) Jun 09 2013 __traits(getAttributes) returns a tuple. Try using an alias instead,

"Temtaime" <temtaime gmail.com> writes:

 (3) int a;
enum tp = __traits(getAttributes, a);
				
enum b = tp[0];
writeln(b);

Compiler doesn't accept enum b = tp[0] and says to me:
Error: variable _tp_field_0 cannot be read at compile time

Why? And writeln(tp[0]) works as expected.

David <d dav1d.de> writes:

Am 09.06.2013 11:15, schrieb Temtaime:
  (3) int a;
 enum tp = __traits(getAttributes, a);
                
 enum b = tp[0];
 writeln(b);
 
 Compiler doesn't accept enum b = tp[0] and says to me:
 Error: variable _tp_field_0 cannot be read at compile time
 
 Why? And writeln(tp[0]) works as expected.

It works like that:
http://dpaste.dzfl.pl/809de989

 (3) int a;
enum tp = [__traits(getAttributes, a)];
enum b = tp[0];
pragma(msg, b);


The exact reason why you have to expand the tuple into an array, I don't
know, or can't remember ;)

"Temtaime" <temtaime gmail.com> writes:

Oh, thanks. But then that code doesn't work:
http://dpaste.dzfl.pl/262b966c

"Temtaime" <temtaime gmail.com> writes:

Thanks for all, it's unactual anymore.

Jacob Carlborg <doob me.com> writes:

On 2013-06-09 11:15, Temtaime wrote:
  (3) int a;
 enum tp = __traits(getAttributes, a);

 enum b = tp[0];
 writeln(b);

 Compiler doesn't accept enum b = tp[0] and says to me:
 Error: variable _tp_field_0 cannot be read at compile time

 Why? And writeln(tp[0]) works as expected.

__traits(getAttributes) returns a tuple. Try using an alias instead, 
something like:

import std.typetuple
alias tp = TypeTuple!(__traits(getAttributes, a));

-- 
/Jacob Carlborg