Paul Tagliamonte <[email protected]> writes: > Here's a minimal example: > > / > | #include <stdio.h> > | #include <nettle/rsa.h> > | > | int main ( int argc, char ** argv ) { > | struct sha1_ctx hash_ctx; > | uint8_t headers[] = { > | 0xDE, 0xEA, 0xAD, > | 0xBE, 0xEF > | }; > | sha1_update(&hash_ctx, 5, headers);
You must call sha1_init(&hash_ctx) before the first sha1_update. > Does doing the sha1_digest alter state in some way? I can't image it is, but > then how does this behavior crop up? sha1_digest implies a sha1_init. So with sha1_digest(...); sha1_digest(...), the second call gives you the sha1 digest of the empty string. I think this is explained in the manual, http://www.lysator.liu.se/~nisse/nettle/nettle.html#Legacy-hash-functions Regards, /Niels -- Niels Möller. PGP-encrypted email is preferred. Keyid C0B98E26. Internet email is subject to wholesale government surveillance. _______________________________________________ nettle-bugs mailing list [email protected] http://lists.lysator.liu.se/mailman/listinfo/nettle-bugs
