Jtdcftul

float ff = (float)31.15;



double dd = 31.15;



var frst = Math.Round(ff, 1, MidpointRounding.AwayFromZero);



var drst = Math.Round(dd, 1, MidpointRounding.AwayFromZero);

frst: 31.1

drst: 31.2

Can someone explain why?

Well, Math.Round wants double, not float, that's why

Math.Round(ff, 1, MidpointRounding.AwayFromZero);

equals to

Math.Round((double)ff, 1, MidpointRounding.AwayFromZero);

and if we inspect (double)ff value

Console.Write(((double)ff).ToString("R"));

we'll see round up errors in action

31.149999618530273

Finally, Math.Round(31.149999618530273, 1, MidpointRounding.AwayFromZero) == 31.1 as expected

In floating point, all numbers are represented internally as fractions where the denominator is a power of 2.

(This is a similar way to how decimals are actually fractions with power-of-10 denominators. So 31.15 is just a way of writing the fraction 3115/100)

In floating point, 31.15 must be represented internally as a binary number. The closest binary fraction is: 1111.1001001100110011001100110011001100110011001100110011001100...repeating

The 1100 recurs (repeats forever), and so the number will be truncated depending on whether it is stored in a double or a float. In a float it is truncated to 24 digits, and in a double to 53.

Exact:  1111.100100110011001100110011001100110011001100110011001100110011001100...forever

Float:  1111.10010011001100110011

Double: 1111.1001001100110011001100110011001100110011001100110

Therefore you can see that the double that this number converts to, is actually slightly larger than the float it converts to. So it is clear that it won't necessarily round to the same number, since it is not the same number to begin with.