3 ms·
The "when done correctly" is a huge caveat, given that what you're replying to does it incorrectly. If X and Y collide, so that SHA-1(X) = SHA-1(Y), then f(SHA-
by atq2119 5y ago
The "when done correctly" is a huge caveat, given that what you're replying to does it incorrectly. If X and Y collide, so that SHA-1(X) = SHA-1(Y), then f(SHA-1(X)) = f(SHA-1(Y)) no matter what f is -- so you get a collision.
Concatenating the outputs of the hash functions fixes that particular issue. But concatenating makes preimage attacks easier.