Hi,
Another API alternative would be to add a version that triggers a lambda
on overflow and allow the caller to determine the value or throw.
public static long multiplyExact(long x,long y,LongBinaryOperator recompute) {
long r =x *y;
long ax =Math.abs(x);
long ay =Math.abs(y);
if (((ax |ay) >>>31 !=0)) {
// Some bits greater than 2^31 that might cause overflow // Check the
result using the divide operator // and check for the special case of
Long.MIN_VALUE * -1 if (((y !=0) && (r /y !=x)) ||
(x ==Long.MIN_VALUE &&y == -1)) {
return recompute.applyAsLong(x,y);
}
}
return r;
}
$.02, Roger
On 12/13/24 8:22 AM, Raffaello Giulietti wrote:
On 2024-12-12 23:58, Charles Oliver Nutter wrote:
Question two: Am I losing the benefits of *Exact if I use the
following code to "pre-check" for overflow?
long high = Math.multiplyHigh(a, b);
if (high == 0) return Math.multiplyExact(a, b);
return bigIntegerMultiply(a, b);
For your specific multiplication use case you might try with
long high = Math.multiplyHigh(a, b);
long low = a * b;
if (high == 0) return low;
return "the big integer consisting of high and low";
It might be possible that the multiplyHigh() and * on the same
operands, when appearing adjacent to each other, get optimized to just
one instruction.
And if not, they might be executed "in parallel" inside the CPU.
HTH
Raffaello
