How Do I Get a Double Result When Dividing Two Integers in C#?

Achieving Double Precision in C# Integer Division

Challenge:

How can we obtain a double-precision floating-point result when dividing two integers in C#?

Solution:

The key is to explicitly cast at least one of the integer operands to a double before performing the division. This forces the division operation to be carried out using floating-point arithmetic, yielding a double result.

<code class="language-csharp">double num3 = (double)num1 / num2; </code>

In this example, num1 is cast to a double. C#'s type system will then promote num2 to a double before the division, ensuring a double-precision outcome. Alternatively, you can cast both:

<code class="language-csharp">double num3 = (double)num1 / (double)num2;</code>

It's important to note that if one or both operands are already doubles, C# automatically performs a double division, producing a double result without explicit casting.

<code class="language-csharp">double num1 = 10.0;
int num2 = 5;
double num3 = num1 / num2; // num3 will be a double</code>

Further Reading:

For a deeper understanding of type casting and floating-point arithmetic in C#, consult online resources such as Dot Net Perls.

The above is the detailed content of How Do I Get a Double Result When Dividing Two Integers in C#?. For more information, please follow other related articles on the PHP Chinese website!