Re: double is integer?

"Joseph M. Newcomer" <newcomer@xxxxxxxxxxxx> wrote in message
news:0dcas2lrmirbjg293b14fd9odq39ide0au@xxxxxxxxxx
This is problematic at best. While it might work, in general it is not a
safe thing to do
because what if your value is 2.0000000000001 or 1.99999999999999? After
doing a lot of
floating point computations, you may have a value close to but not
precisely an integer.

For example
double d = 10.0;
double q = d / 3.0;
double t = q * 3.0;

you might assume that d == t, which will not be true, and isInteger(d) may
be true but
isInteger(t) will not be.

On the whole, this is a very risky thing to consider doing. You should
never assume that
any computation involving a double will ever produce an integer value.
For that matter,
if the integer value is very large, the double might lose low-order
precision anyway.

There is no reliable way to actually compare doubles to integers and know
that the double
is an integer value. Typically, you might consider

if( (d - ceil(d)) < fuzzfactor)

but that is about as good as it gets.
joe


How about:

bool isInteger(const double& d)
{

long int i = (long int) d;
if ((double) i == d)
{
// we have an integer
return true;
}
else
return false;
}


.

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