Why Do Repeated Calls to `Random.Next()` in a Tight Loop Produce Identical Numbers?

Generating Random Numbers Efficiently in Loops

The following code snippet demonstrates a common pitfall when generating random numbers within a loop:

<code class="language-csharp">// Inefficient random number generation
public static int RandomNumber(int min, int max)
{
    Random random = new Random();
    return random.Next(min, max);
}</code>

Calling this function repeatedly in a tight loop, for example:

<code class="language-csharp">byte[] mac = new byte[6];
for (int x = 0; x < 6; x++)
{
    mac[x] = (byte)RandomNumber(0, 255);
}</code>

will often result in identical values within the mac array. This is because Random is re-initialized with the system clock in each iteration. If the loop executes quickly, the clock hasn't changed significantly, resulting in the same seed and thus the same sequence of "random" numbers.

The Solution: A Single, Shared Instance

The correct approach involves creating a single Random instance and reusing it throughout the loop. Thread safety should also be considered for multi-threaded applications. Here's an improved version:

<code class="language-csharp">// Efficient and thread-safe random number generation
private static readonly Random random = new Random();
private static readonly object syncLock = new object();

public static int RandomNumber(int min, int max)
{
    lock (syncLock)
    {
        return random.Next(min, max);
    }
}</code>

By using a static readonly instance, we ensure only one Random object is created. The lock statement protects against race conditions in multi-threaded scenarios, guaranteeing that only one thread accesses the random.Next() method at a time. This maintains the integrity and randomness of the generated numbers.

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