Okay, that is a technical/implementation problem - and I'm not the guy who can answer than question.
I was thinking there might be a mathematical (or other) operator that doesn't work without something in front of it to operate on, and I chose the ^ operator based on a wild guess, plus the following: > php -r "var_dump(4 ^ 8);" int(12) > php -r "var_dump(^ 8);" Parse error: syntax error, unexpected '^', expecting ')' in Command line code on line 1 > php -r "$foo = ^ 8;" Parse error: syntax error, unexpected '^' in Command line code on line 1 I don't know the parser guts, but I figured since using the ^ operator with nothing to operate on, causes a parser-error, that means the parser knows that this is currently *not* valid syntax? Which might mean that there's a way to turn this into valid syntax? I don't know. I also figured the ambiguity with a bitwise operator is minimal damage - bitwise operators are not one of the most commonly used features in high-level languages. If the asterisk (or some other character) offers and easier implementation path, whatever. On Tue, Apr 30, 2013 at 7:55 PM, Rasmus Lerdorf <ras...@lerdorf.com> wrote: > On 04/30/2013 03:24 PM, Rasmus Schultz wrote: > > Are we really going to quibble about syntax? This adds nothing to this > > discussion. And as I explained earlier, the ^ operator is used for the > sake > > of discussion only - if it's more practical to use another character for > > this operator, I don't care what it looks like. > > The point is that there is no operator that fits and the reason you > didn't come up with one that didn't clash with something else. If this > can even be implemented, which doesn't seem all that certain without a > lot of messy changes, it should just be a reflection call. > > -Rasmus >
