Writeup for the hello world challenge:

Description for the challenge:


I am gonna be honest I can't able to get any info from the description

Okk lets proceed to our always goto 3 steps to stage the arena to get the flag for the challenge.

The first 3 steps is normal execution,ltrace and strings.Let's perform these steps and find out any info



Okk from the normal execution step the info we can be able to get is the program checks if 3 user arguements(program_name,2 other user arguements) and after it checks whether the first arguement is correct then it proceeds to check whether the next arguement is correct

And now the ltrace part ...Whattt sort of black magic is this how come ltrace does not produce any sort of output!!Pretty Strangeee!!!!!

Time for analyzing the binary using strings,the information we can able to scratch from here is there are 2 user defined functions:-check and print.According to name, the check function maybe used to check whether the provided arguements are correct and then the print function maybe used to print the flag if both arguements are correct.And the presence of strcmp gives positive signs of string compare if program debugged in gdb we can set breakpoint in the instruction that calls the strcmp function and analyze the rsi and rdi registers and see the values that are compared which makes it pretty easy..Then the name of the source code of the challenge is chall.c

Though we know the presence of strcmp in binary..Still before using gdb let's try ghidra's decompiler and try to get more information about the binary or the flag itself in this phase ...




The main seems to be small and the information we can get from this decompiled version of main is the program first check whether there are 3 user arguements(1 file name arguement(which is default),2 user arguements) and if it is succeded it moves it another if statement that checks whether the return value from check function is equal to 0 when the first arguement for the function is first user arguement and the int 1.And if the return value is zero it checks the next if statement which proceeds the same way as the first if statement but the difference is the arguements which is second arguement and int 2.And if both the if statement succeeds the print function is executed which generates and print the flag.And if the statement does not succeed it gives which arguement is not correct.

Let's move to the check function where it looks little bit complex but can be duplicated because of absence of undefined..But a strcmp is present which gonna make it way easier than that of duplicating the program.. From the decompiled version we can see that the first execution of check function executes first part of check function with strcmp and returns 0 if both values are same and proceeds to the second time execution of check function where the next strcmp check where both values are same and return 0 and if not return values other than zero.

Frankly I have no idea about what is going in the print function.

Armed with this knowledge let's move to gdb and set breakpoints on the call strcmp instruction in check function and examine the values in rsi and rdi register where one holds the value of our user input arguement where the other holds the value by the binary..We have to execute the program three times to get the flag.

First time to get first arguement and the second time to get the value of the second arguement and third time with both correct arguements to get the flag.


Okk we have set breakpoints at both the call strcmp instructions and ready to start execution of the program by two sample flag format arguements.Now let's wait and see the value of the first arguement.


After hitting the breakpoint 1 we found that our first input arguement is compared with the value "H3ll0" which we found out by examining the rdi register and then we didn't hit on breakpoint 2 ..But it's no surprise as we found from the decompiled version the execution passes to second part of check function only if the first arguement is correct.So lets try again the second time with first arguement as H3ll0.


We can see that for first strcmp both value is same and this time we are able to hit the second breakpoint where after examining the rsi and rdi registers we can see that our second input arguement is compared with the value "W0rld" and after continuing the execution the program throws the error that "invalid argument: darkCTF{5678}" showing that second arguement is wrong .Now this time lets change the second arguement to "W0rld" and see what happens this time ..Lets run this time with arguements "H3ll0" "W0rld".


For the first breakpoint the values of rsi and rdi registers are same and for the second breakpoint the same value occurs in both rsi and rdi registers.Continuing the execution of the program ..And Voila!! The flag of the challenge appers.

The flag is darkCTF{4rgum3nts_are_v3ry_1mp0rt4nt!!!}

Original writeup (https://github.com/vital-information-resource-under-siege/DarkCTF-Writeups/blob/master/HelloWorld/README.md).