Creating implementation for idiom 238 : Xor byte arrays

Idiom

# 238 Xor byte arrays

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

a and b have the same size.

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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();