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+# TryHackMe-Intro to x86-x64
+
+Username : zx
+password : reismyfavl33t
+
+SSH into the machine with these credentials.
+
+## Task 2
+
+Executing `intro` binary will produce this result
+
+```
+tryhackme@ip-10-10-213-153:~/introduction$ ./intro
+value for a is 1 and b is 2
+value of a is 2 and b is 1
+
+```
+
+To debug this we use `radare` .
+
+`r2 -d intro`
+
+```
+tryhackme@ip-10-10-213-153:~/introduction$ r2 -d intro
+Process with PID 1507 started...
+= attach 1507 1507
+bin.baddr 0x55ebd6a73000
+Using 0x55ebd6a73000
+asm.bits 64
+ -- Ask not what r2 can do for you - ask what you can do for r2
+[0x7fdc18484090]>
+```
+
+Using command `aa` will scan for all symbols and entry points in the executable.
+
+```
+[0x7fdc18484090]> aa
+[x] Analyze all flags starting with sym. and entry0 (aa)
+[0x7fdc18484090]>
+```
+
+Finding list of functions in `radare` is command `afl` that returns a list of functions in a binary
+
+
+
+Now to analyze a function we use a command `pdf @function_name` which stands for print disassembly function
+
+
+
+
+Here left column instruction which is starting from `0x55ebd6a7366a` are memory addresses of instructions
+
+Middle column are instructions `4883ec08` encoded in bytes which is machine code.
+
+Right column is the human readable instruction `subq $8, %rsp`
+
+%rsp stack pointer (holds the recent memory address)
+%rbp frame pointer (points to frame of function being executed)
+
+
+## Simple Instruction Example
+
+An assembly instruction looks like this
+
+`movq $3 rax ` (movq for 64 bit)
+`movl $3 rax` (movl for 32 bit)
+
+In assembly constatns are represented by this `$3`
+
+Here the instruction will move a value 3 into a register named rax.
+
+To move value from register to register
+
+`movq %rax %rbx`
+
+
+There are some more instructions in assebmly like
+
+```
+addq (add instruction)
+subq (subtract instruction)
+imulq (multiply instruction)
+andq (perform AND operation)
+orq (perform OR operation)
+xorq (perform XOR operation
+```
+
+## If Statements
+
+if (x == 3 ):
+ return 3
+else:
+ return 0
+
+Like we see in programming if else statements that have a conidition that needs to be true and if it's true what instruction it performs or if it's false what will it do so assembly it's something like this
+
+`cmpq %rax %rbx` This will just compare the values of two registers
+
+Now after we compare these values it would not do anything until we tell it what instruction is to be performed so we issue a `jump` instruction which can goto different part of code from a point where that instruction occurs
+
+```
+movq $10 rax
+cmp %rax %rbx
+je equal
+
+equal:
+```
+What this block of instruction will do is assign `rax` register a value of 10
+
+```
+rax = 10
+```
+then will compare `rax` and `rbx` it would not do anything until we perform a jump compare (je equal) if they are equal
+
+```
+if rax == rbx
+
+```
+
+Then it will jump to next block instruction which we have named `equal` it's more like a function
+
+So now let's do some room tasks
+
+
+
+I have loaded up the binary in `radare`
+
+
+
+Then selecting main function and adding two breakpoints with `db [instruction_memory_address]`
+
+
+
+When we'll run the binary it will hit at the breakpoint and two see the instructions uptil that point we'll use `dr` command
+
+
+
+Now for doing these tasks we have to use `if2` binary
+
+
+
+Now add a breakpoint at
+
+`0x55feb072c630 816dfce70300. subl $0x3e7, var_4h`
+
+```
+db 0x55feb072c630 (this is before the return and pop instruction)
+dc
+px @ rbp-0x4
+px @ rbp-0x8
+```
+
+1. What is the value of var_8h before the popq and ret instructions?
+ You'll get in the offset 60 which is in hex so after converting it to decimal it would be
+
+ 
+
+ 
+
+ `96`
+
+2. what is the value of var_ch before the popq and ret instructions?
+
+ We can see that only value 0 is assigned to `var_ch` and no further instructions and done on it until pop and return instruction
+
+ `0`
+
+3. What is the value of var_4h before the popq and ret instructions?
+
+ `1`
+
+4. ` What operator is used to change the value of var_8h, input the symbol as your answer(symbols include +, -, *, /, &, |):`
+
+
+```
+0x55feb072c623 8365f864 andl $0x64, var_8h
+```
+`&`
+
+## Loops
+
+
+
+
+
+
+
+
+
+1. What is the value of var_8h on the second iteration of the loop?
+ `5`
+
+
+2. What is the value of var_ch on the second iteration of the loop?
+ `0`
+
+
+
+3. What is the value of var_8h at the end of the program?
+
+ `2`
+
+4. What is the value of var_ch at the end of the program?
+
+ `0`
+
+
+## Crackme 1
+
+
+
+
+
+Now we know the password is 127 in between you can see `.` and `.01` so by this it is referring to localhost which is 127.0.0.1
+
+`127.0.0.1`
+
+
+## Crackme 2
+
+
+
+Here we see a text file which being read form the binary on reading the file we see `vs3curepwd` but when entered it isn't the right password ,so I got stuck but through "reversing" the text that was the right password
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