How to Get Int31n Type Random Number in Golang?
Last Updated :
26 Mar, 2020
Go language provides inbuilt support for generating random numbers of the specified type with the help of a math/rand package. This package implements pseudo-random number generators. These random numbers are generated by a source and this source produces a deterministic sequence of values every time when the program run. And if you want to random numbers for security-sensitive work, then use the crypto/rand package.
You are allowed to generate a non-negative pseudo-random number in [0, n) of 31-bit integer n as an int32 type from the default source with the help of the Int31n() function provided by the math/rand package. So, you need to add a math/rand package in your program with the help of the import keyword to access the Int31() function. This method will panic if the value of n is <= 0.
Syntax:
func Int31n(n int32) int32
Let us discuss this concept with the help of the given examples:
Example 1:
package main
import (
"fmt"
"math/rand"
)
func main() {
res_1 := rand .Int31n(2)
res_2 := rand .Int31n(4)
res_3 := rand .Int31n(1)
fmt.Println( "Random Number 1: " , res_1)
fmt.Println( "Random Number 2: " , res_2)
fmt.Println( "Random Number 3: " , res_3)
}
|
Output:
Random Number 1: 1
Random Number 2: 3
Random Number 3: 0
Example 2:
package main
import (
"fmt"
"math/rand"
)
func main() {
res_1 := rand .Int31n(0)
res_2 := rand .Int31n(-1)
fmt.Println( "Random Number 1: " , res_1)
fmt.Println( "Random Number 2: " , res_2)
}
|
Output:
panic: invalid argument to Int31n
goroutine 1 [running]:
math/rand.(*Rand).Int31n(0x43e280, 0x0, 0x43e280, 0x460000)
/usr/local/go/src/math/rand/rand.go:128 +0x180
math/rand.Int31n(…)
/usr/local/go/src/math/rand/rand.go:324
main.main()
/tmp/sandbox269134149/prog.go:16 +0x40
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