c# - Why does this method return double.PositiveInfinity not DivideByZeroException? -


i ran following snippet in vs2015 c# interactive , got weird behavior.

> double divide(double a, double b) . { .     try .     { .         return / b; .     } .     catch (dividebyzeroexception exception) .     { .         throw new argumentexception("argument b must non zero.", exception); .     } . }     > divide(3,0) infinity     > 3 / 0 (1,1): error cs0020: division constant 0 > var b = 0; > 3 / b attempted divide zero. >

why did method return infinity while 3 / 0 threw error , 3 / b threw formated error? can force division have thrown error instead of returning infinity?

if reformat method to

double divide(double a, double b) {     if ( b == 0 )     {         throw new argumentexception("argument b must non zero.", new dividebyzeroexception());     }     return / b; }

would new dividebyzeroexception contain same information , structure caught exception would?

it's because use system.double.

as stated msdn dividebyzeroexception thrown integral types , decimal.

that's because hard define "so called" 0 double value.

positiveinfinity results division 0 positive dividend, , negativeinfinity results division 0 negative dividend. (source: msdn again)

dividebyzeroexception not appropriate floating point types. note: can nan though when attempting divide 0 dividend of zero.


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