Creating implementation for idiom 238 : Xor byte arrays

Idiom

# 238 Xor byte arrays

Created by

Lead paragraph

Write in a new byte array c the xor result of byte arrays a and b.

a and b have the same size.

New implementation

Language

Imports

Code

Documentation URL

Original attribution URL

Online runnable demo

Username

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations

#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])

import static java.util.stream.Stream.iterate;

int n = a.length;
byte c[] = new byte[n];
iterate(0, i -> i + 1)
    .limit(n)
    .forEach(i -> {
        c[i] = (byte) (a[i] ^ b[i]);
    });

int i, n = a.length;
byte c[] = new byte[n];
for (i = 0; i < n; ++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();