This assembly is for Intel x86-64 bit, seems to be too baffling to me.
- How come the
jzinstruction jump to a non-instruction (
- How come the
callinvokes a non-existing address?
For curious ones, the binary is here.
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How come the jz instruction jump to a non-instruction (0x400AC9)?
It does not. There is no such thing as "an instruction". Jumps do not jump to instructions, they jump to addresses.
How come the call invokes a non-existing address?
It does not. If you check what the code actually does, you will find this call will never be executed.
Here is the relevant part of your code, prefixed with their original hex bytes.
400AC7: 66 b8 eb 05 mov ax,0x5eb 400ACB: 31 c0 xor eax,eax 400ACD: 74 fa jz 0x400AC9 400ACF: e8 0f b6 45 b0 call 0xffffffffb045b61c 400AD4: 3c 38 cmp al,0x38 400AD6: 0f .byte 0xf 400AD7: 85 .byte 0x85 400AD8: bc .byte 0xbc
If you assume the instruction at 400AC7 executes, it loads
ax with some value but it gets immediately discarded by the next instruction, which clears it to
0. Because of that, the jump will always be taken!
The jump goes to 400AC9, and if we disassemble starting at that, we get some other code:
400AC9: eb 05 jmp 0x400AD0 400ACB: 31 c0 xor eax,eax 400ACD: 74 fa je 0x0 400ACF: e8 0f b6 45 b0 call 0xffffffffb045b61a 400AD4: 3c 38 cmp al,0x38 400AD6: 0f .byte 0xf
and it looks like you did not get any further, because there are still weird calls and undefined codes. However, look at the first instruction! It jumps again into the middle of something what appears to be an instruction at this point – but if you start disassembling there, you will see that everything from that address on is perfectly ordinary code.*
* Since this is a "crackme" and you already failed the first of its tests, I will not spoil the rest of it.