How to Get Int63n 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 63-bit integer n as an int64 type from the default source with the help of the Int63n() 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 Int63n() function. This method will panic if the value of n is <= 0.
Syntax:
func Int63n(n int64) int64
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 .Int63n(7)
res_2 := rand .Int63n(3)
res_3 := rand .Int63n(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: 5
Random Number 2: 1
Random Number 3: 0
Example 2:
package main
import (
"fmt"
"math/rand"
)
func int63nrandom(value_1, value_2 int64) int64 {
return value_1 + value_2 + rand .Int63n(4)
}
func main() {
res1 := int63nrandom(10, 3)
res2 := int63nrandom(430, 56)
res3 := int63nrandom(13, 500)
fmt.Println( "Result 1: " , res1)
fmt.Println( "Result 2: " , res2)
fmt.Println( "Result 3: " , res3)
}
|
Output:
Result 1: 15
Result 2: 489
Result 3: 514
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