Why Does Casting a Float Expression to an Integer in C# Produce Different Results Depending on the Method?

Unusual behavior of C# float to integer conversion

The following code snippet demonstrates unexpected behavior of floating point to integer conversion in C#:

<code class="language-csharp">int speed1 = (int)(6.2f * 10);
float tmp = 6.2f * 10;
int speed2 = (int)tmp;</code>

Surprisingly, speed1 has a value of 61, while speed2 has a value of 62, even though both expect the result to be 62. Close inspection reveals that in speed1 the floating point expression 6.2f * 10 is implicitly converted to type double before being truncated to an integer, resulting in a result of 61.

However, in speed2, the result of 6.2f * 10 is explicitly assigned to the float variable tmp. This intermediate step involves rounding the value to the nearest float value, which is 62, and then converting it to an integer.

The key difference between the two operations is the intermediate representation. In the first case, the compiler might choose to reserve a higher-precision representation for the floating-point expression, causing truncation and yielding a value of 61. However, in the second case, the explicit float assignment forces rounding, resulting in a value of 62.

In the following actions:

<code class="language-csharp">double d = 6.2f * 10;
int tmp2 = (int)d;</code>

d will be of type double, causing double to be converted and the rounded value truncated to an integer, which usually results in a value of 62.

Unexpected behavior in floating-point expression conversions is a subtle aspect of C#'s floating-point handling. Understanding this nuance is critical to avoid unexpected results and ensure accurate conversions between numeric types.

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