Some Assembly Required 3
- 1.The new WebAssembly base64 string is as follows:
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 - 2.Let's decompile it by first converting it to
wasmusingwrite_wasm.pyin../Some Assembly Required 2and then usingwasm-decompile. The decompiled c-like code is in wasm-decompile-output.c. - 3.We can decompile the compiled wasm to actual c code using
wasm2c, which is also included in WebAssembly/wabt.wasm2c(Documentation) creates harder to read long c code that will actually run whilewasm-decompile(Documentation) creates easier to read c-style pseudocode. - 4.We can use
git diffor diffchecker.com to compare the new wasm with the wasm from the previous challenge. We see the addition of a new variable, calledkey, and some changes at the end of thecopy_char. - 5.The encrypted flag is:
0x9d, 0x6e, 0x93, 0xc8, 0xb2, 0xb9, 0x41, 0x8b, 0x9f, 0x90, 0x8c, 0x62, 0xc5, 0xc3, 0x95, 0x88, 0x34, 0xc8, 0x93, 0x92, 0x88, 0x3f, 0xc1, 0x92, 0xc7, 0xdb, 0x3f, 0xc8, 0x9e, 0xc7, 0x89, 0x31, 0xc6, 0xc5, 0xc9, 0x8b, 0x36, 0xc6, 0xc6, 0xc0, 0x90, 0x00, 0x00. This value is from thewasm2coutput file because it shows a clear array of hex bytes while thewasm-decompileoutput file shows a slight mess. - 6.The key is:
0xf1, 0xa7, 0xf0, 0x07, 0xed. - 7.If we analyze the c and c-like code, we will discover that the flag is xored with the reverse of the key. I'm not sure why the key needs to be reversed, but it could have something to do with big endian and little endian since
wasmis close to machine code. - 8.An alternative way to find the key is as follows: We know from step 6 that the key is 5 bytes. We also know that the flag starts with
picoCTF{. Thus, we can xor the start of the flag (picoC) with the encrypted flag0x9d, 0x6e, 0x93, 0xc8, 0xb2to find that they key is:0xed 0x07 0xf0 0xa7 0xf1(CyberChef Recipe). This is the reverse of the key found in the code in step 6. - 9.Now that we have the key and encrypted flag, we can simply decode the encrypted flag as hex and then xor the encrypted flag and key to decrypt the flag (CyberChef Recipe).
picoCTF{8aae5dde384ce815668896d66b8f16a1}