The logic here was that, really, the cast to real should be implicit,
but using implicit casts would be ambiguous, so I created an explicit
sqrt() for every integer type that explicitly casts to real.
On 8/12/2010 3:39 AM, Jonathan M Davis wrote:
On Wednesday 11 August 2010 23:40:17 Daniel Murphy wrote:
Do we really want the sqrt of a integer to return a real?
I would expect it to return floor(sqrt(n)), in the same way that integer
division does.
What do other people think/expect?
That seems rather limiting. Don't you usually want a floating point value from
sqrt? If you want an int, you can just cast the result. Otherwise, you're going
to have to cast the ints that you pass to sqrt() in order to get a floating
value
result, which would tend to be even more annoying - particularly since I would
expect the programmer to want a floating point result normally anyway. On top of
that, because an int value can be held just fine in a floating point, making
sqrt() return an int could easily cause bugs where someone expects it to return
a floating value and assigns it to a floating value. The assignment works just
fine, but they never get the result that they expect, and they may have a hard
time tracking down such a bug.
If it returns a real, then it does what I would expect most people to want, and
if you want an int, you have to cast it. It's nice and clear, and you don't have
errors due to something being a type that you don't expect.
- Jonathan MDavis
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