.NET C# float.Parse returning different result to double.Parse

Can some more experienced guys explain this strange error I found
today?
I was getting strange amounts when I was loading table data with my C#
script.

As it turns out the problem is a
different output from similar
functions:

string
            amount_in_string = "1234567,15";
double amount_in_double =
            double.Parse(amount_in_string);
float amount_in_float =
            float.Parse(amount_in_string);

//amount_in_double =
            1234567.15;
//amount_in_float =
            1234567.13;

Why do I
get such a different result when float and double are similar types (floating point).
Can the precision make a difference with small amounts like
these?

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class="normal">Answer

When
“1234567.15” is converted to double, the result is the closest
value representable in double, which is
1234567.1499999999068677425384521484375. Although you report in the question that the
value is 1234567.15, the actual value is 1234567.1499999999068677425384521484375.
“1234567.15” would be displayed when the value is displayed with a limited number of
decimal digits.

When “1234567.15” is converted
to float, the result is the closet value representable in
float, which is 1234567.125. Although you report the value is
1234567.13, the actual value is 1234567.125. “1234567.13” may be displayed when the
value is displayed with a limited number of decimal
digits.

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Observe that 1234567 exceeds 1,048,576, which is
2²⁰. The 32-bit floating-point format used for
float uses 24 bits for the significand (fraction portion of the
number). If the high bit of the significand represents 2²⁰, the
low bit represents 2²⁰⁻²³ = 2⁻³ = ⅛. This
is why you see “1234567.15” converted to a value rounded to the nearest
eighth.