Why Does Calling `rand.Intn` Without Seeding Result in the Same Sequence of Numbers in Go?

Deterministic Random Number Generation in Go

In Go, calling rand.Intn(n int) int without seeding the random number generator will always return the same sequence of numbers. This is because the default source used for random number generation is initialized with a fixed seed, effectively making it deterministic.

Why Does This Happen?

As mentioned in the official documentation, the rand.Intn function returns a pseudo-random number, which is generated using a deterministic algorithm and a seed value. If the seed is not specified, the default seed of 1 is used, leading to the generation of the same series of numbers in every run.

Properly Seeding Random Number Generation

To generate truly random numbers, it is crucial to seed the random number generator with a different seed for each run. This can be achieved by calling the rand.Seed() function and passing it a random seed value. A common practice is to use the current Unix timestamp as the seed, which ensures that the generator is initialized with a unique value for each execution.

Example:

<code class="go">import (
    "fmt"
    "math/rand"
    "time"
)

func main() {
    rand.Seed(time.Now().UnixNano())
    fmt.Println(rand.Intn(10))
}</code>

Conclusion

By seeding the random number generator properly, you can ensure that your Go programs generate truly random numbers, making them unpredictable and more representative of true randomness.

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