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With the aim to speed up the RSA decryption process, sometimes, "Chinese Remainder Theorem" is used. The theory with the practice about the usage of this theorem may be found in the article "Using the CRT with RSA".


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Now when a compiler runs printf("%s", text);, it would look into the stack and go to the address 5054. However what about printf(text)? How does it access the address of 5054 to print "%s" which after printing might result to a segmentation fault? It doesn't. In the case where the user enters "%s" as the input, there will be two strings at different ...


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I do not see a stack buffer overflow (or any other vulnerability) in the code you have presented.


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using ndisasm, the data in the sh array can be disassembled into the following valid 64bit x86-machinecode: 00000000 EB0B jmp short 0xd 00000002 5F pop rdi 00000003 4831D2 xor rdx,rdx 00000006 52 push rdx 00000007 5E pop rsi 00000008 6A3B push byte +0x3b 0000000A 58 ...



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