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mknos
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Jan 2, 2024
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Maybe the MemCopy() is writing over the ciphertext buffer.
| res |= a[0](U32) << 0 * 8; | ||
| res |= a[1](U32) << 1 * 8; | ||
| res |= a[2](U32) << 2 * 8; | ||
| res |= a[3](U32) << 3 * 8; |
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Does promotion to U32 occur after one byte at a[3] is accessed? Considering << is multiplying by a power of 2, why do you need to multiply again by 8?
| I64 i; | ||
| // Define the plaintext message | ||
| U8 plaintext[16] = "Hello, TempleOS!"; | ||
| U64 plaintext_len = sizeof(plaintext); |
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Did it properly in #188 |
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