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With the Merkle Damgard Transform, how does it handle messages that are shorter than the input message length - hash digest length? Or in other words, how is the last block handled? So for example, if I have a 40 bit long message and I want to use SHA 256 to get a hash digest, how would I do that? Doesn’t SHA 256 require a 512 bit message and the IV plus my message is only 296 bits? Thanks guys

jack klompus
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  • Dupe https://crypto.stackexchange.com/questions/2753/in-the-sha-hash-algorithm-why-is-the-message-always-padded https://crypto.stackexchange.com/questions/1427/why-does-the-padding-in-merkle-damg%C3%A5rd-hash-functions-like-md5-contain-the-messa https://crypto.stackexchange.com/questions/9369/how-is-input-message-for-sha-2-padded https://crypto.stackexchange.com/questions/47149/why-does-sha-2-call-for-doing-10-padding-in-addition-to-appending-the-message-l – dave_thompson_085 Feb 16 '18 at 01:33

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In SHA-256, the message is padded to a multiple of the block length by appending a single 1 bit and then as many 0 bits are needed to fill a message block to 448 bits; finally the 64-bit length is appended to make the input length an exact multiple of 512 bits.

Squeamish Ossifrage
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