go - Want to choose a random number and not have it picked again until all numbers have gone through -

i'm looking pick random number 1-6 if example 1 picked program want make sure not again used until 2-6 has been picked. want in order have chance go through options instead of having same option 2 or maybe 3 times in row since there 6 options. ideas please?

//choose random number recipe rand.seed(time.now().utc().unixnano()) myrand := random(1, 6) fmt.println(myrand) 

...

function processes it

//random number function func random(min, max int) int {     return rand.intn(max-min) + min 

}

the part of program uses myrand each run used in if statement , tied recipe picked day

//choose random number recipe rand.seed(time.now().utc().unixnano()) myrand := random(1, 6) fmt.println(myrand)  //test below of random replacement list := rand.perm(6) i, _ := range list {     fmt.printf("%d\n", list[i]+1) }  //logic recipe choose if myrand == 1 {     fmt.println(1)     printrecipeoftheday(recipe1) } else if myrand == 2 {     fmt.println(2)     printrecipeoftheday(recipe2) } else if myrand == 3 {     fmt.println(3)     printrecipeoftheday(recipe3) } else if myrand == 4 {     fmt.println(4) } 

}

i'm wondering if i'm doing incorrectly ? i'm not sure how can use list on every run generate new number (myrand) , program able remember last number (myrand) when ran previous time?

you can use rand.perm(6) , loop through result.

for example, here code print numbers 1-6 in random order, no repeats:

package main import "fmt" import "math/rand"  func main(){     list := rand.perm(6);     i, _ := range list {         fmt.printf("%d\n", list[i]+1);     } } 

note we've added 1 everything, since perm(6) returns 0-5 , want 1-6. after you've used six, call rand.perm(6) again 6 new numbers.