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I was reading RFC 2898 and something is not clear to me. When I use PBKDF2 with SHA-256 and I want a derived key with length 32 bytes (the same length as my hash function output), your derived key only exists out 1 block (l = 1) of size hLen = 32 bytes.

This seems trivially, but i want to be very sure! Is this correct?

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The output length of the F step of PBKDF, i.e. the T_i each are of the size of your hash function's (or actually: your PRF's) output.

So yes, when the desired output size is as large (or smaller) as the hash function output, we have l = 1 and thus only one call to F(P, S, c, 1).

I suppose this is also the most common way to use PBKDF-2, the extension to longer generated key sizes is only included to make the scheme more generally usable.

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  • $\begingroup$ The extension to longer key sizes is useful when the output is going to be an actual encryption key or other non-password storage use. Storing an output longer than native hash size is actively counterproductive (BAD) when the output is a password hash (since the attacker only needs to attack the first block, and you wasted time calculating more than one, thus giving the attacker a [further] speed advantage over you). $\endgroup$
    – Anti-weakpasswords
    Commented Feb 25, 2014 at 5:46

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