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Decoding and understanding assembly code
I am a beginner with c and assembly code, we have an "bomb" assignment (written in c)which calls methods that require certain passwords, but the code is not visible and I need to determine the correct password by looking at the assembly code.
The code indicates the password for this method is 6 numbers, which is passed as "input" to method puzzle_1 (I am trying to avoid triggering ).
I can't understanding assembly code.
What is the answer to this question?
I think this puzzle_1's keyword is array.
08048db4 <puzzle_1>:
8048db4: push %ebp
8048db5: mov %esp,%ebp
8048db7: sub $0x38,%esp
8048dba: lea -0x24(%ebp),%eax
8048dbd: mov %eax,0x4(%esp)
8048dc1: mov 0x8(%ebp),%eax
8048dc4: mov %eax,(%esp)
8048dc7: call 804897e <read_six_numbers>
8048dcc: movl $0x1,-0xc(%ebp)
8048dd3: jmp 8048df9 <puzzle_1+0x45>
8048dd5: mov -0xc(%ebp),%eax
8048dd8: mov -0x24(%ebp,%eax,4),%eax
8048ddc: mov -0xc(%ebp),%edx
8048ddf: sub $0x1,%edx
8048de2: mov -0x24(%ebp,%edx,4),%edx
8048de6: add $0xbf,%edx
8048dec: cmp %edx,%eax
8048dee: je 8048df5 <puzzle_1+0x41>
8048df0: call 8048d93 <denied_nextstep>
8048df5: addl $0x1,-0xc(%ebp)
8048df9: cmpl $0x5,-0xc(%ebp)
8048dfd: jle 8048dd5 <puzzle_1+0x21>
8048dff: call 8048d73 <allow_nextstep>
8048e04: leave
8048e05: ret