Semantics of ^^, Version 3 (Final?)

[Don]

Rainer Deyke wrote:
> Don wrote:
>> Rainer Deyke wrote:
>>> Not quite.  Under the proposal, -1^^-1 works (i.e. produces the correct
>>> result) at compile time but fails at runtime.
>> It won't pass CTFE.
> 
> pure int f() { return -1; }
> void g(int)(int);
> g!(f() ^^ f())(0); // Works.

No, it doesn't work. It's exactly the same as:

int intpow(int x, int y)
{
    assert(y>=0);
    return x;
}

g!(intpow(f(), f())(0);


f() ^^ f() is not a constant expression. It won't get transformed. It 
can be *interpreted* at compile time, but that'll generate an error.


> g!(0)(f() ^^ f()); // Runtime error?

Definitely.
> 
> 'f() ^^ f()' can be a compile-time constant, but isn't guaranteed to be
> evaluated at compile time.

No, it's not a compile-time constant, unless it's turned into one.

If however you changed it to:

template eval(int x) {  enum int eval = x; }

g!( eval!(f()) ^^ f() )(0);

which turns f() into a compile-time constant, then it'd work.
But then it'd work at runtime, as well.