Semantics of ^^

[Don]

Based on everyone's comments, this is what I have come up with:

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x ^^ y is right associative, and has a precedence intermediate between 
multiplication and unary operators.

* The type of x ^^ y is the same as the type of x * y.
* If y == 0,  x ^^ y is 1.
* If both x and y are integers, and y > 0,  x^^y is equivalent to
    { auto u = x; foreach(i; 1..y) { u *= x; } return u; }
* If both x and y are integers, and y < 0, an integer divide error 
occurs, regardless of the value of x. This error is detected at compile 
time, if possible.
* If either x or y are floating-point, the result is pow(x, y).
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Rationale:
(1) Although the following special cases could be defined...
  * If x == 1,  x ^^ y is 1
  * If x == -1 and y is even, x^^y == 1
  * If x == -1 and y is odd, x^^y == -1
... they are not sufficiently useful to justify the major increase in 
complexity which they introduce. In all other cases, a negative exponent 
indicates an error; it should be rewritten as (cast(real)x) ^^ y. Making 
these cases errors makes everything much simpler, and allows the 
compiler to use range propagation on the value of y to detect most 
exponentiation errors at compile time. (If those cases are legal, the 
compiler can't generate an error on x^^-2, because of the possibility 
that x might be 1 or -1).
Also note that making it an error leaves open the possibility of 
changing it to a non-error later, without breaking code; but going from 
non-error to error would be more difficult.

(2) USE OF THE INTEGER DIVIDE ERROR
Note that on x86 at least, a hardware "integer divide error", although 
commonly referred to as "division by zero", also occurs when the DIV 
instruction, which performs uint = ulong/uint, results in a value 
greater than uint.max. Raising a number to a negative power does involve 
a division, so it seems to me not unreasonable to use it for this case 
as well.
Note that 0 ^^ -1 is a division by zero.
This means that, just as you should check that y!=0 before performing 
x/y, you should check that y>=0 before performing x^^y.

(3) OVERFLOW
int ^^ int returns an int, not a long. Although a long would allow 
representation of larger numbers, even doubling the number of bits 
doesn't help much in avoiding overflow, because x^^y is exponential. 
Even a floating-point representation can easily overflow:
5000^5000 easily overflows an 80-bit real.
So, it's preferable to retain the simplicity that typeof(x^^y) is 
typeof(x*y).