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It's a buffer overflow. They can get a vulnerable implementation to execute arbitrary code, so they can get it to return any value they want. The actual SHA-3 a
by SAI_Peregrinus 4y ago
It's a buffer overflow. They can get a vulnerable implementation to execute arbitrary code, so they can get it to return any value they want. The actual SHA-3 algorithm wouldn't return the same value as the vulnerable implementation, of course.