Given two strings as input, return the result of XORing the code-points of one string against the code points of the other.
For each character in the first input string, take the code-point (e.g. for A, this is 65) and XOR the value against the corresponding index in the second string and output the character at the code-point of the result. If one string is longer than the other, you must return the portion of the string beyond the length of the shorter, as-is. (Alternatively, you may pad the shorter string with NUL bytes, which is equivalent.)
See the following JavaScript code for an example:
const xorStrings = (a, b) => {
let s = '';
// use the longer of the two words to calculate the length of the result
for (let i = 0; i < Math.max(a.length, b.length); i++) {
// append the result of the char from the code-point that results from
// XORing the char codes (or 0 if one string is too short)
s += String.fromCharCode(
(a.charCodeAt(i) || 0) ^ (b.charCodeAt(i) || 0)
);
}
return s;
};
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Test cases
Input Output
['Hello,', 'World!'] '\x1f\x0a\x1e\x00\x0b\x0d'
['Hello', 'wORLD'] '?*> +'
['abcde', '01234'] 'QSQWQ'
['lowercase', "9?' "] 'UPPERCASE'
['test', ''] 'test'
['12345', '98765'] '\x08\x0a\x04\x02\x00' _not_ 111092
['test', 'test'] '\x00\x00\x00\x00'
['123', 'ABCDE'] 'pppDE'
['01', 'qsCDE'] 'ABCDE'
['`c345', 'QQ'] '12345'
Rules
- The two input strings will only ever be code-points 0-255.
- This is code-golf so the shortest solution, in each language, wins.