Write in a new byte array c the xor result of byte arrays a and b.a and b have the same size.
#include <array> #include <cstddef>
std::array<std::byte, a.size()> c; for (auto ia = a.begin(), ib = b.begin(); auto & rc : c) { rc = *ia++ ^ *ib++; }
using System.Linq;
var c = a.Zip(b, (l, r) => (byte)(l ^ r)).ToArray();
use iso_fortran_env, only : int8
integer(kind=int8), dimension(:) :: a, b, c ! Assign values to a and b c = ieor(a,b)
c := make([]byte, len(a)) for i := range a { c[i] = a[i] ^ b[i] }
var c T for i := range a { c[i] = a[i] ^ b[i] }
const c = Uint8Array.from(a, (v, i) => v ^ b[i])
byte[] c = new byte[a.length]; for (int i = 0; i < a.length; i++) { c[i] = (byte) (a[i] ^ b[i]); }
(map '(vector (unsigned-byte 8)) #'logxor a b)
local c = {} for i=1,#a do c[i] = string.char(string.byte(a, i) ~ string.byte(b, i)) end c = table.concat(c)
SetLength(c, Length(a)); for i := Low(a) to High(a) do c[i] := a[i] xor b[i];
use feature 'bitwise';
$c = $a ^. $b;
c = bytes([aa ^ bb for aa, bb in zip(a, b)])
c = a.zip(b).map{|aa, bb| aa ^ bb}
let c: Vec<_> = a.iter().zip(b).map(|(x, y)| x ^ y).collect();
No security, no password. Other people might choose the same nickname.