RC4 Encryption KSA key out of bounds

Question

I am implementing a RC4 algorithm in C# for my WPF app.

I followed this pdf stating the algorithm.

The thing is, in the KSA, we ought to do : j = (j + S[i] + (int)key[i % keyLengthInBits]) % 256;.

That being said, I do not understand how this would work because the key ought to be from 5 characters long to 32 characters long (40bits up to 256bits).

So let's take an example with a 5 characters key ( I'll use the same as in the pdf I linked above) : pwd12. You go fetch the character of the key at the position i % 40 (5 characters long is 40 bits). It's fine for the first 5 times because from i = 0 to i=4, we have a value in the key (pwd12). Though here is the problem (from how I see it) : when we are at i=5, we do not have any characters left in the key. Therefore we will get an ÒutOfBounds Exception.

How is it possible to works if we try to fetch a character in the key where there is none? Obviously there is something I do not see in the algorithm because it does work otherwise it would not be used...

Return the remainder the division. Therefore it's not the keyLength in bits but the key length itself. Thanks for opening my eyes! — Marks
– Marks, Commented May 16, 2016 at 8:13
It is the key length in bits. The purpose of that expression is to select a single bit of the key. See my answer. — David Schwartz
– David Schwartz, Commented May 16, 2016 at 9:45

David Schwartz · Accepted Answer · 2016-05-16 09:43:54Z

1

The phrase key[i % keyLengthInBits] is a way of saying "the i'th bit of the key". It does not mean the i'th entry in the key expressed as an array of bytes, each holding 8 bits of the key.

The equivalent C code would be something like:

int bit = i & keyLengthInBits; ((key[bit / 8] >> (bit % 8)) & 1)

answered May 16, 2016 at 9:43

David Schwartz

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7 Comments

Marks Over a year ago

Ok then ((key[bit / 8] >> (bit % 8)) & 1) ,which I am not quite sure with, would be that : From what I understand of it, let's say we are at i=5 and the key is still pwd12 (40bits). it would be bit = 5 & 40 which would give 0 and therefore key[0/8] = 'p'. From what I get of the >> operator, it does a right shift on the bit of 0 since (bit % 8 = 0) so p stays from 01110000. The &1 makes it become 00000000. (From here I'm lost about why we would make it become 0...) I guess I'm wrong but from what I understand, this is how it goes in your two lines. @David Schwartz

David Schwartz Over a year ago

@Marks You have it right. If i is 5, the code extracts the fifth bit of the key.

Marks Over a year ago

Then what if we are at the 52th iteration ( i = 52 ). It goes 52 (00110100) & 40 (00101000) which gives 00100000 so bit = 32. key[32/8] = key[4] which gives 2 (still considering the key being pwd12). Right shift of 0 since 32 % 8 = 0. So it ends up being 00000010 & 00000001 which gives 0. Therefore here again it would become j = (j + S[k] + (int)key[0]) % 256; ) being the result we have from the i % ketLengthInBits using the two lines you provided. Is that correct ? @David Schwartz

David Schwartz Over a year ago

If key[4] is 00000010, then as i goes from 32 to 39, it reads off the 32nd to 39th bits from right to left, outputting 0, 1, 0, 0, 0, 0, 0, 0.

Marks Over a year ago

Reading from right to left is because of the >> right shift I'm assuming ? Even though it does not shift since it's >> 0, it does read it from the right. Considering that, we still would go to 0 doing the &1 since 01000000 & 00000001 gives 00000000. No ? @David Schwartz

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