Re: double is integer?

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

On Sat, 3 Feb 2007 16:44:06 +0100, "Guido Franzke" <guidof73@xxxxxxxx> wrote:

Hello NG,

the following source let's me check if double is integer:

bool isInteger(const double& d)
{
if ( d == (double)((int)d) ) return true;
return false;
}


isInteger(2.); -> should return true
isInteger(2.5); -> should return false
isInteger(2.001); -> should return false

My question: Is this ok, or will I get problems? Is there another (secure)
possibility?

TIA
Guido

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