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  • For loop from assembly to C

    - by FranXh
    I have a bomb project where I need to defuse certain phases by finding "pas phrases" that will defuse the bomb. Right now I have been working with phase_2, for which the assembly code is shown below. Phase_2 requires as an input 6 numbers, which I need to find in order to defuse this phase. I analyzed this assembly, and I came up with the C code below, that covers lines from 40101c to 401044. It is basically a for loop that makes sure that elements t[0]==t[3], t[1]==t[4] and t[2]==t[5] that the user inputs, are equal. According to my logic, the user can input any 6 numbers as long as the above condition is satisfied. Say 1, 2, 3, 1, 2, 3 would be a valid "pas phrase". However this solution does not convince me for some reason. Am I doing something wrong? 0000000000400ffc <phase_2>: 400ffc: 48 89 5c 24 e0 mov %rbx,-0x20(%rsp) 401001: 48 89 6c 24 e8 mov %rbp,-0x18(%rsp) 401006: 4c 89 64 24 f0 mov %r12,-0x10(%rsp) 40100b: 4c 89 6c 24 f8 mov %r13,-0x8(%rsp) 401010: 48 83 ec 48 sub $0x48,%rsp 401014: 48 89 e6 mov %rsp,%rsi 401017: e8 65 0a 00 00 callq 401a81 <read_six_numbers> 40101c: 48 89 e5 mov %rsp,%rbp 40101f: 4c 8d 6c 24 0c lea 0xc(%rsp),%r13 401024: 41 bc 00 00 00 00 mov $0x0,%r12d 40102a: 48 89 eb mov %rbp,%rbx 40102d: 8b 45 0c mov 0xc(%rbp),%eax 401030: 39 45 00 cmp %eax,0x0(%rbp) 401033: 74 05 je 40103a <phase_2+0x3e> 401035: e8 2d 09 00 00 callq 401967 <_GLOBAL_RESET_> 40103a: 44 03 23 add (%rbx),%r12d 40103d: 48 83 c5 04 add $0x4,%rbp 401041: 4c 39 ed cmp %r13,%rbp 401044: 75 e4 jne 40102a <phase_2+0x2e> 401046: 45 85 e4 test %r12d,%r12d 401049: 75 05 jne 401050 <phase_2+0x54> 40104b: e8 17 09 00 00 callq 401967 <_GLOBAL_RESET_> 401050: 48 8b 5c 24 28 mov 0x28(%rsp),%rbx 401055: 48 8b 6c 24 30 mov 0x30(%rsp),%rbp 40105a: 4c 8b 64 24 38 mov 0x38(%rsp),%r12 40105f: 4c 8b 6c 24 40 mov 0x40(%rsp),%r13 401064: 48 83 c4 48 add $0x48,%rsp 401068: c3 for (int i=0; i<3; i++){ if(t[i] != t[i+3]){ explode(); } }

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  • Do "if" statements affect in the time complexity analysis?

    - by FranXh
    According to my analysis, the running time of this algorithm should be N2, because each of the loops goes once through all the elements. I am not sure whether the presence of the if statement changes the time complexity? for(int i=0; i<N; i++){ for(int j=1; j<N; j++){ System.out.println("Yayyyy"); if(i<=j){ System.out.println("Yayyy not"); } } }

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