Re: integer abs() overflow
Jonathan Lee wrote:
On Jul 5, 4:37 pm, Pete Becker <p...@versatilecoding.com> wrote:
No, the code that uses it exhibits the same problem. <g>
Then I misunderstood you. I thought you were suggesting that it would
do the conversion for me.
Yes, sorry I wasn't clear. The problem is that for some representations
there is no conversion.
Instead you were suggesting that it would
allow me to detect values that were out of range. I suppose then I
could do
unsigned long my_abs(long a) {
unsigned long c = 0;
long const m = std::numeric_limits<long>::max();
long b = std::abs(a);
while (b < 0) { // move b into range for std::abs()
c += static_cast<unsigned long>(m);
b += m;
b = std::abs(b);
}
c += static_cast<unsigned long>(b);
return c;
}
I guess I would still have to check for c overflowing.
I thought there would be something simpler.
If your world consists of four values, -2, -1, 0, and 1, there's no
value that corresponds to the absolute value of -2. No sequence of
additions and subtractions will get around that. The problem in 2's
complement representations is the same: the smallest negative value has
no corresponding positive value, so you cannot meaningfully compute its
absolute value. The simplest approach to computing an absolute value
simply negates negative values, and for that smallest value the result
is the original negative value. If you're worried about that, you can do
one of two things: don't call abs with that value, or check afterwards
whether you still have a negative value. Either way, you have to decide
what your program should do if this happens. Think of it like dividing
by zero: if you do it, you've got a logic error.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of
"The Standard C++ Library Extensions: a Tutorial and Reference"
(www.petebecker.com/tr1book)