
1 Feb
2008
1 Feb
'08
12:36 p.m.
Thanks for your answer. I am sorry, I guess my shuffling through the proto documentation hasn't be thorough enough ! I'll probably look a bit further into how you can place literals in your expressions "just like that", but it definitely gives me a solution for this matter.
As far as i understand what you have done, this implies instantiating both an expr<> type and the associated context at the same time. Sorry, I don't understand. I was under the wrong impression that expr<> type would not hold any data (only type information) and that contexts existed to compensate for that. I was clearly wrong as shown in the example you've given.
Thanks again, i definitely will look more into your work on Proto !