You cant see me picoctf
We put together a bunch of resources to help you out on our website! If you go over there, you might even find a flag! Can you find the flag in file? This would be really obnoxious to look through by hand, see if you can find a faster way. Using netcat nc will be a necessity throughout your adventure. Can you connect to shell1.
SEE VIDEO BY TOPIC: Piping Stdin & Redirecting Stderr - in out error  picoCTF 2018Content:
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The challenge was worth points, so in the mid to upper range of difficulty. My solution is a little unconventional since I didn't use the provided hints but it still led me to the right solution.
It's pretty easy to spot what's going on. The vuln function uses gets to read our input, so we can easily write as much data as we want to the stack, thus overwriting EIP. This function in turn has three conditions that have to be satisfied. The global variables win1, win2, win3 must be set to true.
Fair enough. When we first did the challenge, PIE was actually enabled. This was a mistake by the challenge author and made the challenge even harder. Fun fact that error was later corrected by the author. For this writeup, let's assume you joined later and wanted to solve it without PIE.
At the end, we will take a brief look how to defeat PIE. We can quickly rule out spawning a shell as the binary does not contain any int 0x80 gadgets. We could try to leak some Libc-addresses and jump to system directly, but that's more work than it's worth. Lets look at the disassembly of leap3 and see if the cat is still alive:. We can directly jump to 0x , assuming that EAX somehow contains the correct address to load the right offset.
Which is a hell of an assumption anyway. But this lead's us to another problem. At the end of win3, there is an mov ebx, [ebp - 0x4] instruction followed by leave; ret;. But in the function we overflow, the saved ebp is stored directly above eip. So overwriting the stored instruction pointer, means overwriting the stored base pointer. No matter what, we can already predict that our exploit will outright destroy the stack management, like when the Undertaker threw Mankind off Hell In A Cell, and plummeted 16 ft through an announcer's table.
And since we would need to jump to at least 4 different places, we will encounter a stack pivot. Who am I kidding? The only thing we will encounter is a good old segfault. This means we can't just straight up supply our whole chain at once, because we would need to supply new input to a "new stack"at some point. Already getting some headache even thinking about our ropchain, we looked back at the display flag function to see if there is any way we can solve it without setting any win flags.
This is the disassembly of the display flag function We are looking for ropgadgets, remember? Nothing special. The flag gets loaded first.
As soon as we call the function, the flag get's loaded onto the stack, and not zeroed out after the return. The win conditions only determine wether it is printed. Since we have to do a stack pivot anyway, why not just use this to load the flag at a known address and use puts to print it?
From here on, we will continue to work directly on the exploit script, using the following skeleton:. This will just crash the program. Also note: I'm using tilix as an emulator, if you don't: replace line 6 by something fitting to your setup, or just delete it :.
We have a special challenge in that we have to move the stack and still supply new input. As soon as the stack points into another data segment, the next return would try to read an address from there. So we can't just use leave; ret; , instead we have to:. Luckily we can reuse parts of the code for the last two tasks. Looking at vuln again, following the code from address 0xb8 we can see that EBP gets dereferenced, and pushed onto the stack. Afterwards gets is called.
And since we jumped into the middle of a function, we even get a free leave; ret; at the end. The third one seems perfect. So, where do we pivot our stack? Well, you can use any not ASLR affected writable data segment, e.
You can find it either by looking in gdb during run-time:. We just need to remember that the stack grows upwards against lower addresses. So we need our fake stack to start at a higher address, otherwise you will get random segfaults.
Lets add Stage 1, the stack pivot, to our script:. Running this until we hit gets and looking at the debugger shows the following:. This means our input will get loaded to that address. Also, and this is important to pivot, EBP now points to 0xa93c. So now the program wants new input. To understand what's going on, just type ni in gdb to continue, and input a bunch of 'A' into the program. I chose to input 28 'A's and 4 'B's ;. Now type ni until you hit leave and observe the stack before.
After leave your gdb should look like this:. ESP now points to 0xa Why does it point directly to our 4 B's and not the start of our input string? Well, that's for you to figure out ; But this means, any data that get's stored on the stack in some way, will now be at a known offset from our fake stack.
This means we could for example leak addresses to libc. Or any other data Now, as we had originally planned, lets add a second stage to load our flag on the stack to our script:.
Running this in gdb until the end of display flag is shown below. Then we would need to return to our next gadget, but that one of course we haven't figured out yet so lets ignore that for now. As we can see ESP has advanced or gone back, because stack another 4 bytes. Our EBP is obiously broken because we hit another leave , which loaded our last 4 'A's of the stage2 payload into EBP, but we don't care at this point.
Just remember: If we encounter another leave , it will destroy our stack as it will try to load this value into ESP. Anyway, the point was to get the flag.
I just created a txt file in the same folder named flag. The heap is pretty useless for us sine ASLR is enabled on the server and we can't predict this address at all. But the flag on our fake stack is pretty much all we need. You can run the program over and over again, it will always be at the same offset. The real stack will of course be at another address every time you run it, but the program itself will always be located at the same address.
This means, we can just write this address down in our exploit script and accept it as a given. The last step is to get that flag out of memory and in our terminal. We have two functions conveniently linked in our binary: puts and printf. If you step through printf you will notice it needs an absurd amount of memory on the stack. And since our stack-size is pretty limited, let's just go with puts. We can use any puts call we want. But due to our stack magic you have to be careful what argument actually get's passed to puts.
As an excersise use puts GOT instead of the call. See what argument is on top of the stack and if you can get the flag address there. I recommend adding a breakpoint at puts and taking a look what's on top. To keep things simple, let's just reuse some of the program code, e. This will just print whatever the top of the stack points to. Now it would be possible to get a clean exit, but then we would have to be more careful about registers like EBP.
So meh. Putting things together in our script:. I advise you to step through puts. Take a look at the stack addresses as it works it's magic and try to figure it out on your own. When we call puts we are in the last third of the diagram. Now puts want's its own stack frame of course. This means, it will overwrite our flag on the stack. Now we have two options: Move the flag, or move the stack No writeup without a dope rhyme btw.
Moving the flag means getting the heap pointer somehow and printing from heap. Anyone here who want's to do this? Yeah, didn't think so.
The challenge was worth points, so in the mid to upper range of difficulty. My solution is a little unconventional since I didn't use the provided hints but it still led me to the right solution. It's pretty easy to spot what's going on.
It may be aimed for younger students but as I am still learning reverse engineering it was perfect for me. Your mission is to enter Dr. The laboratory is protected by a series of locked vault doors. Each door is controlled by a computer and requires a password to open.
PicoCTF 2018 Writeup: General Skills
I'm collecting solutions to small tasks in the "Miscellaneous" category here, since they are too short to warrant individual posts. We put together a bunch of resources to help you out on our website! If you go over there, you might even find a flag! Can you find the flag in file? This would be really obnoxious to look through by hand, see if you can find a faster way. Using netcat nc will be a necessity throughout your adventure. Can you connect to shell3. Just run nc shell3. Can you find the flag in this file without actually running it? The task name is also its solution : run strings on the file and filter the output with grep to get the flag :.
PicoCTF 2018 Write-up for problems 46 through 50
You can learn more about base 2 numbers over here. Other than doing this challenge by hand, you can also solve it using python:. Base 16, base 2, and base 10 are the three most important base in computer science and CTF problems. You can solve this problem both by hand or using python:.
This was a bit of a problem as Linux treats the. As this character has special meaning when it comes to the linux file systems, when you try to cat it normally, you get the error saying that the period character is a directory. This is the advertising link. To be successful on your mission, you must be able read data represented in different ways, such as hexadecimal or binary.
[picoctf 2019] Leapfrog
Stacey Faucett. This is the fifth article in a series of blog posts that describe how Cisco enables nonprofits to maximize technology for greater scale and impact. We live in a world that is more connected than ever before. Connectivity leads to many benefits, such as being able to share ideas and collaborate, access educational content, purchase products online, and access banking more easily.
Get the admin's password. When you first visit the linked webpage you're presented with this:. From the looks of it, a pretty regular login page for what , by the look of it, is supposed to look like your average corporate in-house application. By knowing it's a php server, using the file parameter to render specific pages, is a great start and should help us in doing some additional enumeration. Let's start by trying to figure out some of the available pages by just guessing common filenames for web pages and enter them as the value for the file key. If we enter foot or head , the response is just an empty page, which means our "attack" actually does something and that indeed, these files seem to exist.
PicoCTF 2019: Cereal Hacker 2 (500p)
In recent decades there has been incredible growth in the use of various internet applications by individuals and organizations who store sensitive information online on different servers. This greater reliance of organizations and individuals on internet technologies and applications increases the threat space and poses several challenges for implementing and maintaining cybersecurity practices. Constructing an Ethical Hacking Knowledge Base for Threat Awareness and Prevention provides innovative insights into how an ethical hacking knowledge base can be used for testing and improving the network and system security posture of an organization. It is critical for each individual and institute to learn hacking tools and techniques that are used by dangerous hackers in tandem with forming a team of ethical hacking professionals to test their systems effectively. Highlighting topics including cyber operations, server security, and network statistics, this publication is designed for technical experts, students, academicians, government officials, and industry professionals. IGI Global , 14 dec. Linux Essentials Before We Start.