On Thursday, 28 March 2013 at 21:16:32 UTC, Chris Cain wrote:
I am completely confused as to why you're doing what you are
doing ... std.conv does work (and in the case you've listed, is
unnecessary anyway). Try this:
import std.stdio, std.conv;
void main() {
ubyte[] buffer;
buffer ~= 5; // Simple solution
buffer ~= to!ubyte(6); // Proper usage of "to"
writeln(buffer);
}
This is not what I'm trying to achieve.
This gives me an array with two elements, [5, 6]. What I want is
to append the 4 bytes that make up one integer value, which using
your values means buffer should hold a total of 8 bytes (two
integers).
H. S. Teoh answered well on how this can be achieved, although my
feedback was not really meant as a question of "how is this
done?", more of "why is this done like this, couldn't it be done
much easier?".
