Stack eip overflow x86 vs x86_64 easy C code?

On x86_64, the instruction pointer is RIP not EIP thus if you query the EIP register in gdb with a 64-bit executable, you're not going to get any values since that's not a valid 64-bit register. If you are wanting to keep your executable as 32-bit on a native 64-bit platform, then pass gcc the -m32 flag at compile-time.

On x86_64, the instruction pointer is RIP, not EIP ... thus if you query the EIP register in gdb with a 64-bit executable, you're not going to get any values since that's not a valid 64-bit register. If you are wanting to keep your executable as 32-bit on a native 64-bit platform, then pass gcc the -m32 flag at compile-time. If you are wanting to see how the x86_64 Unix stack behaves compared to the x86 Unix stack, then I suggest reading the x86_64 Unix ABI, sections 3.2 and 3.4.

It's not that there's no overflow on x86_64, it's just a different way of presenting exception. Instead of actually telling you that it was unable to execute the code 0x4141414141414141 after it actually updated rip, it's telling you that the destination address is invalid before updating. This is an architectural difference between x86 code and x86_64 code and whenever you're executing a 64bit code, that's how it is handled.

Again, you'd have different a different message on 64bit code only. You'd have to compile it as 32bit code. There's no way to get the same message if you were to compile it as x86_64 code.It's actually not difficult to notice this difference if you were to properly debug the code and see where rip is pointing to and the values of other registers.

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