Reverse int golang - indexing

Question

How to change 12345 to 54321?
With a string, you can change the string to a rune, and reverse it, but you cannot do the same for an integer. I have searched and found no one talking about this. Examples
131415 >>> 514131
1357 >>> 7531
123a >>> ERROR
-EDIT-
I was thinking, why not create a slice and index that?
Then I realized that you can't index int
(http://play.golang.org/p/SUSg04tZsc)
MY NEW QUESTION IS
How do you index an int?
OR
How do you reverse a int?

Answer 1

Here is a solution that does not use indexing an int
package main
import (
"fmt"
)
func reverse_int(n int) int {
new_int := 0
for n > 0 {
remainder := n % 10
new_int *= 10
new_int += remainder
n /= 10
}
return new_int
}
func main() {
fmt.Println(reverse_int(123456))
fmt.Println(reverse_int(100))
fmt.Println(reverse_int(1001))
fmt.Println(reverse_int(131415))
fmt.Println(reverse_int(1357))
}
Result:
654321
1
1001
514131
7531
Go playground

Answer 2

I converted the integer to a string, reverse the string, and convert the result back to a string.
package main
import (
"fmt"
"strconv"
)
func main() {
fmt.Println(reverse_int(123456))
fmt.Println(reverse_int(100))
fmt.Println(reverse_int(1001))
fmt.Println(reverse_int(131415))
fmt.Println(reverse_int(1357))
}
func reverse_int(value int) int {
intString := strconv.Itoa(value)
newString := ""
for x := len(intString); x > 0; x-- {
newString += string(intString[x - 1])
}
newInt, err := strconv.Atoi(newString)
if(err != nil){
fmt.Println("Error converting string to int")
}
return newInt
}

Answer 3

Very similar to the first answer but this checks to make sure you don't go out of bounds on the type.
func reverse(x int) int {
rev := 0
for x != 0 {
pop := x % 10
x /= 10
if rev > math.MaxInt32/10 || (rev == math.MaxInt32 /10 && pop > 7) {
return 0
}
if rev < math.MinInt32/10 || (rev == math.MinInt32/10 && pop < -8) {
return 0
}
rev = rev * 10 + pop
}
return rev
}

Answer 4

Also flips negative numbers int
func Abs(x int) int {
if x < 0 {
return -x
}
return x
}
func reverse_int(n int) int {
newInt := 0
sign := 1
if n < 0 {
sign = -1
}
n = Abs(n)
for n > 0 {
remainder := n % 10
newInt = newInt*10 + remainder
n /= 10
}
return newInt * sign
}
func main() {
fmt.Println(reverse_int(-100))
fmt.Println(reverse_int(-1001))
fmt.Println(reverse_int(131415))
fmt.Println(reverse_int(1357))
}

Answer 5

Similar to Fokiruna's answer but also checks for a 32bit overflow
func reverse(x int) int {
result, sign := 0, 1
if(x < 0) {
sign = -1
x = -x
}
for x > 0 {
remainder := x % 10;
result = result * 10 + remainder
x = x/10
}
var checkInt int = int(int32(result))
if checkInt != result {
return 0
}
return result * sign
}