What's the best way to compare Double and Int?

问题:

The following code in C# doesn't work:

int iValue = 0;
double dValue = 0.0;

bool isEqual = iValue.Equals(dValue);

So, the question: what's the best way to compare Double and Int?

回答1:

You really can't compare floating point and integral values in a naive way; particularly, since there's the classic floating point representation challenges. What you can do is subtract one from the other and see if the difference between them is less than some precision you care about, like so:

int iValue = 0;
double dValue = 0.0;

var diff = Math.Abs(dvalue - iValue);
if( diff < 0.0000001 ) // need some min threshold to compare floating points
   return true; // items equal

You really have to define for yourself what equality means to you. For example, you may want a floating point value to round towards the nearest integer, so that 3.999999981 will be "equal" to 4. Or you may want to truncate the value, so it would effectively be 3. It all depends on what you're trying to achieve.

EDIT: Note that i chose 0.0000001 as an example threshold value ... you need to decide for yourself what precision is sufficient for comparison. Just realize you need to be within the normal representational bounds of double which I believe is defined as Double.Epsilon.

回答2:

It's an exceedingly bad idea to compare integers and floating-point numbers for equality in any language. It works for very simple cases, but after you do any math at all, the likliehood of the program doing what you want it to decreases dramatically.

It has to do with the way floating-point numbers are stored on a binary, digital system.

If you are very sure you want to use this, create a class to make you own number with fractions. use one int to maintain the whole number, and another int to maintain the fraction.

回答3:

This really depends on what you consider "equal". If you want your comparison to return true if and only if the double precisely matches the integer value (i.e. has no fractional component), you should cast your int to a double to do the comparison:

bool isEqual = (double)iValue == dValue;

If something like 1.1 would be considered equal to 1, you can either cast the double to an int (if you want to ignore the fractional component altogether) or round the double if you want say 1.9 to equal 2.

回答4:

double val1 = 0;
double val2 = 0.0;
if((val1 - Double.Epsilon) < 0)
{
    // Put your code here
}

      OR

if((val2 - Double.Epsilon) < 0)
{
    // Put your code here
}

where Double.Epsilon is lowest possible value for Double.

回答5:

Nowadays, just about the only time one should be comparing values of types double and either integer or long for strict equality is when, for some reason, one is stuck storing or passing integral quantities as floating-point values and later needs to convert them back. Such conversion may in most cases be most easily accomplished by casting the integral type to double, and then comparing the result of that cast. Note that conversion from long to double may be imprecise if the number is outside the range ±2⁵². Nonetheless, in the days before 64-bit long became available, double was a handy storage type for integer quantities which were too big for a 32-bit int but small enough to be handled by double.

Note that converting a long to double and then doing the comparison will yield an "equal" result if the nominal value of the double doesn't precisely match the long value, but represents the closest possible double to that value. This behavior makes sense if one recognizes that floating-point types don't actually represent a single precise value, but rather a range of values.