Why does this method return double.PositiveInfinity not DivideByZeroException?

Question

I ran the following snippet in the VS2015 C# interactive and got some very weird behavior.

> double divide(double a, double b)
. {
.     try
.     {
.         return a / b;
.     }
.     catch (DivideByZeroException exception)
.     {
.         throw new ArgumentException("Argument b must be non zero.", exception);
.     }
. }    
> divide(3,0)
Infinity    
> 3 / 0
(1,1): error CS0020: Division by constant zero
> var b = 0;
> 3 / b
Attempted to divide by zero.
> 

Why did the method return infinity while 3 / 0 threw an error and 3 / b threw a formated error? Can I force the division to have thrown an error instead of returning infinity?

If I reformat the method to

double divide(double a, double b)
{
    if ( b == 0 )
    {
        throw new ArgumentException("Argument b must be non zero.", new DivideByZeroException());
    }
    return a / b;
}

would the new DivideByZeroException contain all the same information and structure that the caught exception would?

Answer 1

It's because you use System.Double.

As stated by MSDN DivideByZeroException is thrown only for integral types and Decimal.

That's because it is hard to define "so called" zero for Double value.

PositiveInfinity also results from a division by zero with a positive dividend, and NegativeInfinity results from a division by zero with a negative dividend. (source: MSDN again)

DivideByZeroException is not appropriate for floating point types. Note: You can get NaN though when attempting to divide by zero with a dividend of zero.