#advanced
[Step-12] step read repeat
1. Step to next instruction
Let's step to next instruction "mov eax, dword [local_8h]" with "ds" ("local_8h" i.e. "5" is moved to eax)
2. Read general purpose registers
Print gpr using "dr" and confirm value of rax is indeed "5" 0x00000005
3. Repeat steps 1 and 2
Running "ds" once again should now execute "add eax, edx" ~ adds rax "5" and rdx "4" stores the sum in rax
Using "dr", we can see rax is "9" 0x00000009 as expected
4. End
Before executing next instruction "mov dword [local_4h], eax", print the variable "local_4h" using "px @rbp-0x4"
We get "00"
Next instruction is "mov dword [local_4h], eax"
Execute it using "ds" and run "px @rbp-0x4" again, we get "09" expected as value of rax was 0x00000009
Queries & suggestions @Revnexbot
[Step-12] step read repeat
1. Step to next instruction
Let's step to next instruction "mov eax, dword [local_8h]" with "ds" ("local_8h" i.e. "5" is moved to eax)
2. Read general purpose registers
Print gpr using "dr" and confirm value of rax is indeed "5" 0x00000005
3. Repeat steps 1 and 2
Running "ds" once again should now execute "add eax, edx" ~ adds rax "5" and rdx "4" stores the sum in rax
Using "dr", we can see rax is "9" 0x00000009 as expected
4. End
Before executing next instruction "mov dword [local_4h], eax", print the variable "local_4h" using "px @rbp-0x4"
We get "00"
Next instruction is "mov dword [local_4h], eax"
Execute it using "ds" and run "px @rbp-0x4" again, we get "09" expected as value of rax was 0x00000009
Queries & suggestions @Revnexbot
#advanced
[Step-13] "end" ?
Recall that the binary is actually a program that adds up two variables "a" and "b" and stores the result in another variable "c" where "a" and "b" are constants "4" and "5" respectively
Explanation
To get an insight of actual program structure, we reversed "run" using radare
Source code breakdown
Integer a "local_ch" is declared
"var int local_ch @ rbp-0xc"
Similarly, variables b "local_8h" and c "local_4h" are declared
Variable a is assigned constant 4 "mov dword [local_ch], 4" and b is assigned constant "5"
To add them together, we utilised the unused registers "eax" and "edx"
Variables a and b are added as edx and eax respectively "add eax, edx", and result is assigned to variable c "local_4h" afterwards "mov dword [local_4h], eax"
Queries & suggestions @Revnexbot
[Step-13] "end" ?
Recall that the binary is actually a program that adds up two variables "a" and "b" and stores the result in another variable "c" where "a" and "b" are constants "4" and "5" respectively
Explanation
To get an insight of actual program structure, we reversed "run" using radare
Source code breakdown
Integer a "local_ch" is declared
"var int local_ch @ rbp-0xc"
Similarly, variables b "local_8h" and c "local_4h" are declared
Variable a is assigned constant 4 "mov dword [local_ch], 4" and b is assigned constant "5"
To add them together, we utilised the unused registers "eax" and "edx"
Variables a and b are added as edx and eax respectively "add eax, edx", and result is assigned to variable c "local_4h" afterwards "mov dword [local_4h], eax"
Queries & suggestions @Revnexbot
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100k users in less than a month, and now — "It's a good bye👋🏻"
Are you a hacker?
It's a message out there for those who can support us —
We need your help!
If you're a hacker, professional coder or someone who can support us —
1. Integrate "last seen" feature
2. Evade bot searches being detected and blocked by truecaller
3. Buy us some proxies
Please connect to us @Revnexbot