I have some arrays of different struct, and I need to change them by the same function func foo(arr interface{}).
And I use the function in this way foo(&arrayToChange)
Then I find that, I cannot change the array by pointer a simple example for you.
package main
import (
"fmt"
)
func A(out interface{}) {
arr := make([]interface{}, 0)
arr = append(arr, "foo", 2.2)
out = &arr
B(out)
}
func B(out interface{}) {
arr := make([]interface{}, 0)
arr = append(arr, "bar", "foo", "anything")
out = &arr
}
func main() {
arr := make([]interface{}, 0)
arr = append(arr, 1, 2, 3)
fmt.Printf("%T
", &arr)
A(&arr)
fmt.Println(arr)
}
I guess your question there is, even though you do out = &arr inside the function, how come arr in the caller is unchanged.
out is a local variable in your function. You pass to the function &arr, so the value of out is the address of arr. And then you change the value of out to something else. This affects nothing outside the scope of this function. Reassigning the values of local variables never affects anything outside the scope of a function.
It seems what you're trying to do is something like this:
*out = arr
That is, change the value where out is pointing.
package main
import (
"fmt"
)
func A(out *[]interface{}) {
arr := make([]interface{}, 0)
arr = append(arr, 1, 2)
*out = arr
}
func main() {
arr := make([]interface{}, 0)
arr = append(arr, 1, 2, 3)
fmt.Printf("%T
", &arr)
A(&arr)
fmt.Println(arr)
}