如何使用golang实现slowEqual

I tried to implement a slowEqual with golang,but xor operation is limited to int and int8 and I have no idea to convert string to int[] or int8[] , even it can be converted it seems little awkward, and I found bytes.Equal but it seems not a slowEqual implementation.Any advices? This is my impletation.

//TODO real slow equal
func slowEquals(a, b string) bool {
    al := len(a)
    bl := len(b)
    aInts := make([]int, al)
    bInts := make([]int, bl)
    for i := 0; i < al; i++ {
        aInts[i] = int(a[i])
    }
    for i := 0; i < bl; i++ {
        bInts[i] = int(b[i])
    }
    var diff uint8 = uint8(al ^ bl)
    for i := 0; i < al && i < bl; i++ {
        diff |= a[i] ^ b[i]
    }
    return diff == 0
    //长度相等为0
    /*
        abytes := []int8()
        bbytes := []int8()
        al := len(a)
        bl := len(b)
        diff := al ^ bl
        for i := 0; i < al && i < bl; i++ {
            diff |= a[i] ^ b[i]
        }
        return diff == 0
    */
}

Or:(after first answer)

import "crypto/subtle"

func SlowEquals(a, b string) bool {
    if len(a) != len(b) {
        return subtle.ConstantTimeCompare([]byte(a), make([]byte,len(a))) == 1
    }else{
        return subtle.ConstantTimeCompare([]byte(a), []byte(b)) == 1       
    }
}

Perhaps this:

import "crypto/subtle"

func SlowEquals(a, b string) bool {
    if len(a) != len(b) {
        return false
    }
    return subtle.ConstantTimeCompare([]byte(a), []byte(b)) == 1
}

This returns quickly if the lengths are different, but there's a timing attack against the original code that reveals the length of a, so I think this isn't worse.