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Code Golf

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Required fields*

Required fields*

XOR two strings

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;
};

Try it online!

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 so the shortest solution, in each language, wins.

Answer*

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  • \$\begingroup\$ Welcome. You can use $args[1]|% t*y instead [char[]]($args[1]). The % t*y means "call the method toCharArray. And you can omit $i=0. See also Tips for golfing in PowerShell \$\endgroup\$ Commented Jul 19, 2020 at 10:44
  • \$\begingroup\$ Am I wrong or does this not work for two strings of unequal length? \$\endgroup\$ Commented Jul 26, 2020 at 15:51
  • \$\begingroup\$ @Thomas yep. You can see the two arguments are different length. \$\endgroup\$ Commented Jul 26, 2020 at 15:54
  • \$\begingroup\$ 1. You have to quote your input, if it contains spaces. You are actually processing I and am instead of I am fat and You are thin. 2. As you can see in your result, it contains only 1 byte instead of two bytes (am has two bytes), because you process up to the length of the first string. In conclusion, you just omit bytes if the first string is the shorter one and your code crashes if the second string is shorter. \$\endgroup\$ Commented Jul 26, 2020 at 16:10

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