Why Does Casting a C# Float Expression to an Int Produce Unexpected Results?

C# Floating Point Expressions: Unexpected behavior in float to integer conversion

In C#, unexpected behavior may occur when casting the result of a floating point expression to an integer. Consider the following code:

int speed1 = (int)(6.2f * 10);
float tmp = 6.2f * 10;
int speed2 = (int)tmp;

Intuitively, speed1 and speed2 should contain the same value. However, the actual result is:

<code>speed1 = 61
speed2 = 62</code>

This anomaly arises from the order of operations performed. In speed1, the result of the multiplication (6.2f * 10) is immediately converted to an integer, resulting in a truncated value of 61.

In contrast, in speed2, the multiplication result is assigned to tmp as a floating point number, triggering rounding during the implicit conversion to floating point. This rounded value of 62 is then converted to an integer, resulting in the correct result.

To understand this behavior, it is important to realize that floating point numbers (such as 6.2f) are not exact mathematical values, but approximations. Therefore, arithmetic operations on floating point numbers may result in loss of precision due to rounding.

In the case of (int)(6.2f 10) , the compiler may (optionally) preserve the precision of 6.2f 10 even though the result is formally a floating point number. This allows for more precise results. However, if this optimization is not implemented, the expression will be cast to a double-precision floating point number, possibly resulting in a loss of precision.

In contrast, the assignment to tmp = 6.2f * 10 explicitly forces rounding before integer conversion. This ensures consistent results across different systems.

Therefore, to avoid such exceptions, it is generally recommended to use an explicit rounding function, such as Math.Round, when converting floating point numbers to integers.

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