Flipping a slice with a for loop logic error - go

Question

So I am trying to write a method that takes two slices, flips both of them and then gives them to each other.
Ex.
s1 = {1,2,3,4,5}
s2 = {6,7,8,9,10}
Should return:
s1 = {10,9,8,7,6}
s2 = {5,4,3,2,1}
Here is my code:
package main
import(
"fmt"
)
func main(){
f:= [5]int{1,2,3,4,5}
h:= [5]int{6,7,8,9,10}
var sliceF []int = f[0:5]
var sliceH []int = h[0:5]
fmt.Println(reverseReverse(sliceF,sliceH))
}
func reverseReverse(first []int, second []int) ([]int, []int){
//creating temp arrays to hold the traversed arrays before swapping.
var tempArr1 []int = first
var tempArr2 []int = second
//count is used for counting up the tempArrays in the correct order in the For loops
var count int= 0
//goes through the first array and sets the values starting from the end equal to the temp array
//which increases normally from left to right.
for i :=len(first)-1; i>=0;i--{
tempArr1[count] = first[i]
fmt.Println(i)
count++
}
count =0
//same as first for loop just on the second array
for i :=len(second)-1; i>=0;i--{
tempArr2[count] = second[i]
count++
}
//trying to replace the values of the param arrays to be equal to the temp arrays
first=tempArr2
second = tempArr1
//returning the arrays
return first,second
}
When run here is the output:
4
3
2
1
0
[10 9 8 9 10]
[5 4 3 4 5]
*Not I included a print statement in the for loop to check if the index is decreasing properly.
I understand there are better ways to do this but for proof of concept I want to use a for loop.
Any help appreciated. I am new to go and tend to have java habits so I assume somehow my problem is related to that.

Answer 1

This can be done much simpler by realizing there's no need to actually swap the individual elements. Instead, reverse each array and swap their order. Much simpler!
func reverseReverse( a, b []int ) ([]int, []int) {
return reverse(b), reverse(a)
}
func reverse( a []int ) []int {
end := len(a) - 1
// Allocate a new array slice of the same length to copy to.
ret := make( []int, len(a) )
// Copy each element of a into ret, reversed.
for i := range a {
ret[end-i] = a[i]
}
return ret
}
With that revelation, there's little need for the very specialized reverseReverse function. Swap the order yourself.
fmt.Println(reverse(sliceH), reverse(sliceF))
Note that if you just want to take a slice of an array, it's sufficient to write sliceH []int := h[:] without specifying the start and end. The start is assumed to be 0 and the end is the end. Also note there's no need to declare the type, := takes care of that for you.
Even better, you can declare and initialize them directly.
sliceF:= []int{1,2,3,4,5}
sliceH:= []int{6,7,8,9,10}

Answer 2

Short answer:
tempArr1[count] = first[i]
This line is logically identical to:
first[count] = first[i]
Detailed answer:
x := [5]int{} and x := []int{} are in fact two very different assignments. In the first case x is actually a static array. In the second case x is a slice which is in fact a data structure which has a length, capacity and a pointer to the underlying array. Therefore, var tempArr1 []int = first means copy the pointer to the underlying array of first into the tempArr1, so any modification to first[i] will be reflected in tempArr1 and vice versa

Answer 3

For example,
package main
import "fmt"
func reverse(s []int) []int {
for i := 0; i < len(s)/2; i++ {
s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i]
}
return s
}
func main() {
s1, s2 := []int{1, 2, 3, 4, 5}, []int{6, 7, 8, 9, 10}
fmt.Println(s1, s2)
s1, s2 = reverse(s2), reverse(s1)
fmt.Println(s1, s2)
}
Output:
[1 2 3 4 5] [6 7 8 9 10]
[10 9 8 7 6] [5 4 3 2 1]