Semantics of ^^, Version 3 (Final?)

[Simen kjaeraas]

On Wed, 09 Dec 2009 09:36:56 +0100, Don <nospam at nospam.com> wrote:

> CHANGES BASED ON FURTHER COMMENTS
> --------------------
> x ^^ y is right associative, and has a precedence intermediate between
>   unary and postfix operators.
> The type of x ^^ y is the same as the type of x * y.
>
> * If either x or y are floating-point, the result is pow(x, y).
>
> If both x and y are integers, the following rules apply:
>
>   * If x is the compile-time constant 0, x^^y is rewritten as (y==0)? 1  
> : 0
>   * If x is the compile-time constant 1, x^^y is rewritten as (y,1)
>   * If x is the compile-time constant -1 and y is an integer, x^^y is  
> rewritten as (y & 1) ? -1 : 1.
>
>   * If y == 0, x ^^ y is 1.
>   * If y > 0,  x ^^ y is functionally equivalent to
>      { auto u = x; foreach(i; 1..y) { u *= x; } return u; }
>   * If y < 0, an integer divide error occurs, regardless of the value of  
> x.
>
> -----------
> Note that by definining the 0,1, -1 cases as "rewriting" rules rather  
> than return values, it should be clearer that they don't apply to  
> variables having those values.
> I think this covers everything useful, while avoiding nasty surprises  
> like
>
> double y = x ^^ -1; // looks like reciprocal, but isn't!
> // Yes, this IS the same problem you get with double y = 1/x.
> // But that's doesn't make it acceptable. I have a possible solution to  
> that one, too.
>
> I don't think we can afford to spend much more time on this.
> Is everyone happy now?

Might I enquire as to why the exponent should be allowed to be signed
for integer exponentiation? It's been covered several times - it makes
no sense.

Apart from that - great job!

-- 
Simen