


Why Does `i = 10 * 0.69;` in C# Result in 6.8999999999999995 Instead of 6.9?
.NET Floating-Point Arithmetic and Precision Issues
In C#, the statement i = 10 * 0.69;
surprisingly results in 6.8999999999999995
instead of the expected 6.9
. This stems from the inherent limitations of floating-point number representation and arithmetic within the .NET framework.
The misconception is that decimal values like 0.69 can be perfectly represented in binary. However, floating-point numbers (like double
and float
) use a binary system, and many decimal fractions cannot be precisely represented in binary just as 1/3 cannot be exactly represented with a finite number of decimal digits. The compiler calculates and stores the result as a constant during compilation, but this stored value is an approximation of the true mathematical result.
Several solutions exist to address this precision issue:
-
Employ the
decimal
data type:decimal
offers higher precision and can accurately represent6.9
. However, usingdecimal
might lead to slightly reduced performance compared todouble
orfloat
. -
Implement rounding: Functions like
Math.Round()
or string formatting withNumberFormatInfo
allow you to round the result to the desired level of precision. - Value adjustment: You could slightly adjust the input values to compensate for potential rounding errors. This approach, however, adds complexity and might introduce new inaccuracies.
Understanding the limitations of floating-point arithmetic is essential for writing robust and accurate numerical code. By utilizing the appropriate techniques, developers can minimize the impact of these precision limitations and ensure reliable mathematical operations.
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