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I always have a question about how to calculate the stretch of the stack. For example, when I have more than 8 parameters in arm64, he actually uses the area of my previous function call stack. After BL enters the function, he uses SP to add back to get the parameters, which is equivalent to crossing a stack. How can he avoid polluting the previous stack in this case? Thank you for your answer
You are correct: the function arguments which do not fit in registers will be pushed onto the stack before calling your function. Therefore, they will be at addresses with positive offsets from SP on entry to your function, and I can see why you might be concerned that it is not safe to access this memory. However, this memory is in fact "yours".
The ARM Procedure Call Standard section 6.4.2 states "A callee is permitted to modify any stack space used for receiving parameter values from the caller". So, there is no need to worry. The caller is expecting you to access this memory, and even to modify it if you want, and nothing will break if you do.
IDA is able to read the machine instructions of a subroutine and display the relative offsets of each variable that gets stored on the stack.
I'm writing a program that analyzes stack memory, and I would like to be able to put the values stored in the stack memory into their respective variable types. What logic is going on under the hood that enables IDA to display the stack variable offsets?
Thanks for your time.
It infers that from the function's code by looking on where and how stack addresses are used. Like, loading into a 4-byte register and doing arithmetics probably mean that stack memory from which the load was performed belongs to a some int variable.
If you want details on IDA's algorithm, I doubt you can found it. You can look at Avast's Retargetable Decompiler open source project, that performs analysis much like IDA and study its code.
I am now trying to learn ruby-debug gem, but there are many jargons I am unable to catch up. Wondering if anyone could help with the explanations?
I couldn't find them in http://bashdb.sourceforge.net/ruby-debug.html either. The author assumed we already understand them (where can I learn about them anyway?).
For example here is a result of calling help frame in rdb. I helplessly don't understand all the items I bolded.
Move the current frame to the
specified frame number.
A negative number indicates position
from the other end. So 'frame -1'
moves to the oldest frame, and 'frame
0' moves to the newest frame.
Without an argument, the command
prints the current stack frame. Since
the current position is redisplayed,
it may trigger a resyncronization if
there is a front end also watching
over things.
If a thread number is given then we
set the context for evaluating
expressions to that frame of that
thread.
It's not Ruby-specific jargon; it's common to most all debugging.
Regarding stack frames
You've likely seen stack traces:
/usr/local/rvm/gems/ree-1.8.7-2010.02/gems/redgreen-1.2.2/lib/redgreen.rb:28:in `write': Broken pipe (Errno::EPIPE)
from /usr/local/rvm/gems/ree-1.8.7-2010.02/gems/redgreen-1.2.2/lib/redgreen.rb:28:in `output_single'
from /usr/local/rvm/rubies/ree-1.8.7-2010.02/lib/ruby/1.8/test/unit/ui/console/testrunner.rb:72:in `add_fault'
from /usr/local/rvm/rubies/ree-1.8.7-2010.02/lib/ruby/1.8/test/unit/ui/console/testrunner.rb:70:in `to_proc'
from /usr/local/rvm/rubies/ree-1.8.7-2010.02/lib/ruby/1.8/test/unit/util/observable.rb:78:in `call'
The full trace shows you the "call stack". The line at the top is where the exception was thrown, and the lines under it show the path through your code that the program took to get to that point. Each of those lines is a level in the stack, called a "stack frame". So, when an exception is thrown, the current stack frame is the top of the stack. If you move to frame -1 then you're moving to the bottom of the call stack. Think of the call stack like a stack of plates. When you call a function, you add a plate to the stack, and when you return out of that function, you remove a plate from the stack. Each plate is a frame. Since you usually end up calling functions within functions within functions, you end up with fairly deep call stacks, and walking up and down them can be useful in debugging, to evaluate local variables and state at each point in the call stack.
If you'd like to read up more on call stacks, Wikipedia has a nice article.
Regarding threads
Most all modern programming languages are multi-threaded, meaning that they can execute multiple code paths (almost) concurrently. So, imagine for example that you have a visual app, and you perform some expensive computation. Your GUI won't be able to react to any user input while that computation is running, which makes the application appear to be frozen to the user. You would solve this by running two threads: One thread would be responsible for accepting and handling user input and painting the GUI, and the other thread would be responsible for doing your heavy computation. Your compute thread could be stuck in an expensive loop and your GUI thread would keep running and painting the GUI.
If you are running a multi-threaded application, then you you have to select which thread you want to evaluate your debug commands (expressions) in, since each thread will be in different points of your code, and will have different call stacks and different local variables and state and such. This is the evaluation context.
However, I noticed that this is a Rails question, and Rails is (by default) single-threaded, so you shouldn't need to worry about threads.
Chris Heald gave a really fantastic answer. A couple of very small comments. Although Ruby Rails may be single-threaded (by default), depending on which kind of web server you run Ruby/Rails the overall program (webserver + Ruby/Rails) may no longer be single-threaded.
The comment:
Since the current position is
redisplayed, it may trigger a
resyncronization if there is a front
end also watching over things.
There are some debugging front ends that parse output looking for a source-code position in the output so that the front (such as the text editor GNU/Emacs) can show you where you were you are. (For GNU/Emacs it shows this in another editor window.) When you change to a different frame, that front needs to update the display showing where you are.
Although I know this to be the case for the Emacs and the oldish debugger front end ddd, I imagine you have the same thing going on if you are debugging from vim.
I've heard of stackless languages. However I don't have any idea how such a language would be implemented. Can someone explain?
The modern operating systems we have (Windows, Linux) operate with what I call the "big stack model". And that model is wrong, sometimes, and motivates the need for "stackless" languages.
The "big stack model" assumes that a compiled program will allocate "stack frames" for function calls in a contiguous region of memory, using machine instructions to adjust registers containing the stack pointer (and optional stack frame pointer) very rapidly. This leads to fast function call/return, at the price of having a large, contiguous region for the stack. Because 99.99% of all programs run under these modern OSes work well with the big stack model, the compilers, loaders, and even the OS "know" about this stack area.
One common problem all such applications have is, "how big should my stack be?". With memory being dirt cheap, mostly what happens is that a large chunk is set aside for the stack (MS defaults to 1Mb), and typical application call structure never gets anywhere near to using it up. But if an application does use it all up, it dies with an illegal memory reference ("I'm sorry Dave, I can't do that"), by virtue of reaching off the end of its stack.
Most so-called called "stackless" languages aren't really stackless. They just don't use the contiguous stack provided by these systems. What they do instead is allocate a stack frame from the heap on each function call. The cost per function call goes up somewhat; if functions are typically complex, or the language is interpretive, this additional cost is insignificant. (One can also determine call DAGs in the program call graph and allocate a heap segment to cover the entire DAG; this way you get both heap allocation and the speed of classic big-stack function calls for all calls inside the call DAG).
There are several reasons for using heap allocation for stack frames:
If the program does deep recursion dependent on the specific problem it is solving,
it is very hard to preallocate a "big stack" area in advance because the needed size isn't known. One can awkwardly arrange function calls to check to see if there's enough stack left, and if not, reallocate a bigger chunk, copy the old stack and readjust all the pointers into the stack; that's so awkward that I don't know of any implementations.
Allocating stack frames means the application never has to say its sorry until there's
literally no allocatable memory left.
The program forks subtasks. Each subtask requires its own stack, and therefore can't use the one "big stack" provided. So, one needs to allocate stacks for each subtask. If you have thousands of possible subtasks, you might now need thousands of "big stacks", and the memory demand suddenly gets ridiculous. Allocating stack frames solves this problem. Often the subtask "stacks" refer back to the parent tasks to implement lexical scoping; as subtasks fork, a tree of "substacks" is created called a "cactus stack".
Your language has continuations. These require that the data in lexical scope visible to the current function somehow be preserved for later reuse. This can be implemented by copying parent stack frames, climbing up the cactus stack, and proceeding.
The PARLANSE programming language I implemented does 1) and 2). I'm working on 3). It is amusing to note that PARLANSE allocates stack frames from a very fast-access heap-per-thread; it costs typically 4 machine instructions. The current implementation is x86 based, and the allocated frame is placed in the x86 EBP/ESP register much like other conventional x86 based language implementations. So it does use the hardware "contiguous stack" (including pushing and poppping) just in chunks. It also generates "frame local" subroutine calls the don't switch stacks for lots of generated utility code where the stack demand is known in advance.
Stackless Python still has a Python stack (though it may have tail call optimization and other call frame merging tricks), but it is completely divorced from the C stack of the interpreter.
Haskell (as commonly implemented) does not have a call stack; evaluation is based on graph reduction.
There is a nice article about the language framework Parrot. Parrot does not use the stack for calling and this article explains the technique a bit.
In the stackless environments I'm more or less familiar with (Turing machine, assembly, and Brainfuck), it's common to implement your own stack. There is nothing fundamental about having a stack built into the language.
In the most practical of these, assembly, you just choose a region of memory available to you, set the stack register to point to the bottom, then increment or decrement to implement your pushes and pops.
EDIT: I know some architectures have dedicated stacks, but they aren't necessary.
Call me ancient, but I can remember when the FORTRAN standards and COBOL did not support recursive calls, and therefore didn't require a stack. Indeed, I recall the implementations for CDC 6000 series machines where there wasn't a stack, and FORTRAN would do strange things if you tried to call a subroutine recursively.
For the record, instead of a call-stack, the CDC 6000 series instruction set used the RJ instruction to call a subroutine. This saved the current PC value at the call target location and then branches to the location following it. At the end, a subroutine would perform an indirect jump to the call target location. That reloaded saved PC, effectively returning to the caller.
Obviously, that does not work with recursive calls. (And my recollection is that the CDC FORTRAN IV compiler would generate broken code if you did attempt recursion ...)
There is an easy to understand description of continuations on this article: http://www.defmacro.org/ramblings/fp.html
Continuations are something you can pass into a function in a stack-based language, but which can also be used by a language's own semantics to make it "stackless". Of course the stack is still there, but as Ira Baxter described, it's not one big contiguous segment.
Say you wanted to implement stackless C. The first thing to realize is that this doesn't need a stack:
a == b
But, does this?
isequal(a, b) { return a == b; }
No. Because a smart compiler will inline calls to isequal, turning them into a == b. So, why not just inline everything? Sure, you will generate more code but if getting rid of the stack is worth it to you then this is easy with a small tradeoff.
What about recursion? No problem. A tail-recursive function like:
bang(x) { return x == 1 ? 1 : x * bang(x-1); }
Can still be inlined, because really it's just a for loop in disguise:
bang(x) {
for(int i = x; i >=1; i--) x *= x-1;
return x;
}
In theory a really smart compiler could figure that out for you. But a less-smart one could still flatten it as a goto:
ax = x;
NOTDONE:
if(ax > 1) {
x = x*(--ax);
goto NOTDONE;
}
There is one case where you have to make a small trade off. This can't be inlined:
fib(n) { return n <= 2 ? n : fib(n-1) + fib(n-2); }
Stackless C simply cannot do this. Are you giving up a lot? Not really. This is something normal C can't do well very either. If you don't believe me just call fib(1000) and see what happens to your precious computer.
Please feel free to correct me if I'm wrong, but I would think that allocating memory on the heap for each function call frame would cause extreme memory thrashing. The operating system does after all have to manage this memory. I would think that the way to avoid this memory thrashing would be a cache for call frames. So if you need a cache anyway, we might as well make it contigous in memory and call it a stack.
How does a stack overflow occur and what are the ways to make sure it doesn't happen, or ways to prevent one?
Stack
A stack, in this context, is the last in, first out buffer you place data while your program runs. Last in, first out (LIFO) means that the last thing you put in is always the first thing you get back out - if you push 2 items on the stack, 'A' and then 'B', then the first thing you pop off the stack will be 'B', and the next thing is 'A'.
When you call a function in your code, the next instruction after the function call is stored on the stack, and any storage space that might be overwritten by the function call. The function you call might use up more stack for its own local variables. When it's done, it frees up the local variable stack space it used, then returns to the previous function.
Stack overflow
A stack overflow is when you've used up more memory for the stack than your program was supposed to use. In embedded systems you might only have 256 bytes for the stack, and if each function takes up 32 bytes then you can only have function calls 8 deep - function 1 calls function 2 who calls function 3 who calls function 4 .... who calls function 8 who calls function 9, but function 9 overwrites memory outside the stack. This might overwrite memory, code, etc.
Many programmers make this mistake by calling function A that then calls function B, that then calls function C, that then calls function A. It might work most of the time, but just once the wrong input will cause it to go in that circle forever until the computer recognizes that the stack is overblown.
Recursive functions are also a cause for this, but if you're writing recursively (ie, your function calls itself) then you need to be aware of this and use static/global variables to prevent infinite recursion.
Generally, the OS and the programming language you're using manage the stack, and it's out of your hands. You should look at your call graph (a tree structure that shows from your main what each function calls) to see how deep your function calls go, and to detect cycles and recursion that are not intended. Intentional cycles and recursion need to be artificially checked to error out if they call each other too many times.
Beyond good programming practices, static and dynamic testing, there's not much you can do on these high level systems.
Embedded systems
In the embedded world, especially in high reliability code (automotive, aircraft, space) you do extensive code reviews and checking, but you also do the following:
Disallow recursion and cycles - enforced by policy and testing
Keep code and stack far apart (code in flash, stack in RAM, and never the twain shall meet)
Place guard bands around the stack - empty area of memory that you fill with a magic number (usually a software interrupt instruction, but there are many options here), and hundreds or thousands of times a second you look at the guard bands to make sure they haven't been overwritten.
Use memory protection (ie, no execute on the stack, no read or write just outside the stack)
Interrupts don't call secondary functions - they set flags, copy data, and let the application take care of processing it (otherwise you might get 8 deep in your function call tree, have an interrupt, and then go out another few functions inside the interrupt, causing the blowout). You have several call trees - one for the main processes, and one for each interrupt. If your interrupts can interrupt each other... well, there be dragons...
High-level languages and systems
But in high level languages run on operating systems:
Reduce your local variable storage (local variables are stored on the stack - although compilers are pretty smart about this and will sometimes put big locals on the heap if your call tree is shallow)
Avoid or strictly limit recursion
Don't break your programs up too far into smaller and smaller functions - even without counting local variables each function call consumes as much as 64 bytes on the stack (32 bit processor, saving half the CPU registers, flags, etc)
Keep your call tree shallow (similar to the above statement)
Web servers
It depends on the 'sandbox' you have whether you can control or even see the stack. Chances are good you can treat web servers as you would any other high level language and operating system - it's largely out of your hands, but check the language and server stack you're using. It is possible to blow the stack on your SQL server, for instance.
A stack overflow in real code occurs very rarely. Most situations in which it occurs are recursions where the termination has been forgotten. It might however rarely occur in highly nested structures, e.g. particularly large XML documents. The only real help here is to refactor the code to use an explicit stack object instead of the call stack.
Most people will tell you that a stack overflow occurs with recursion without an exit path - while mostly true, if you work with big enough data structures, even a proper recursion exit path won't help you.
Some options in this case:
Breadth-first search
Tail recursion, .Net-specific great blog post (sorry, 32-bit .Net)
Infinite recursion is a common way to get a stack overflow error. To prevent - always make sure there's an exit path that will be hit. :-)
Another way to get a stack overflow (in C/C++, at least) is to declare some enormous variable on the stack.
char hugeArray[100000000];
That'll do it.
Aside from the form of stack overflow that you get from a direct recursion (eg Fibonacci(1000000)), a more subtle form of it that I have experienced many times is an indirect recursion, where a function calls another function, which calls another, and then one of those functions calls the first one again.
This can commonly occur in functions that are called in response to events but which themselves may generate new events, for example:
void WindowSizeChanged(Size& newsize) {
// override window size to constrain width
newSize.width=200;
ResizeWindow(newSize);
}
In this case the call to ResizeWindow may cause the WindowSizeChanged() callback to be triggered again, which calls ResizeWindow again, until you run out of stack. In situations like these you often need to defer responding to the event until the stack frame has returned, eg by posting a message.
Usually a stack overflow is the result of an infinite recursive call (given the usual amount of memory in standard computers nowadays).
When you make a call to a method, function or procedure the "standard" way or making the call consists on:
Pushing the return direction for the call into the stack(that's the next sentence after the call)
Usually the space for the return value get reserved into the stack
Pushing each parameter into the stack (the order diverges and depends on each compiler, also some of them are sometimes stored on the CPU registers for performance improvements)
Making the actual call.
So, usually this takes a few bytes depeding on the number and type of the parameters as well as the machine architecture.
You'll see then that if you start making recursive calls the stack will begin to grow. Now, stack is usually reserved in memory in such a way that it grows in opposite direction to the heap so, given a big number of calls without "coming back" the stack begins to get full.
Now, on older times stack overflow could occur simply because you exausted all available memory, just like that. With the virtual memory model (up to 4GB on a X86 system) that was out of the scope so usually, if you get an stack overflow error, look for an infinite recursive call.
I have recreated the stack overflow issue while getting a most common Fibonacci number i.e. 1, 1, 2, 3, 5..... so calculation for fib(1) = 1 or fib(3) = 2.. fib(n) = ??.
for n, let say we will interested - what if n = 100,000 then what will be the corresponding Fibonacci number ??
The one loop approach is as below -
package com.company.dynamicProgramming;
import java.math.BigInteger;
public class FibonacciByBigDecimal {
public static void main(String ...args) {
int n = 100000;
BigInteger[] fibOfnS = new BigInteger[n + 1];
System.out.println("fibonacci of "+ n + " is : " + fibByLoop(n));
}
static BigInteger fibByLoop(int n){
if(n==1 || n==2 ){
return BigInteger.ONE;
}
BigInteger fib = BigInteger.ONE;
BigInteger fip = BigInteger.ONE;
for (int i = 3; i <= n; i++){
BigInteger p = fib;
fib = fib.add(fip);
fip = p;
}
return fib;
}
}
this quite straight forward and result is -
fibonacci of 100000 is : 25974069347221724166155034021275915414880485386517696584724770703952534543511273686265556772836716744754637587223074432111638399473875091030965697382188304493052287638531334921353026792789567010512765782716356080730505322002432331143839865161378272381247774537783372999162146340500546698603908627509966393664092118901252719601721050603003505868940285581036751176582513683774386849364134573388343651587754253719124105003321959913300622043630352137565254218239986908485563740801792517616293917549634585586163007628199160811098365263529954406942842065710460449038056471363460330005208522777075544467947237090309790190148604328468198579610159510018506082649192345873133991501339199323631023018641725364771362664750801339824312317034314529641817900511879573167668349799016820118499077566864568450662873924856039140476051995500662888263458771894106803700918793650017330117100283104739474562560914449328213748555738640805798130282666402703542944121049199958031318768058991865134251759599115205631553377039969410355182752749199598022575079020377981030899229849963044962558140455170002502997643221934621653662108418767454282982613982344783665815880408190033073829395000821320093747154851310272208173054322648669496309879147143629255542526240439996153269798768075106468190687921182991679644091782718685617029181022126792674013626504997849688436809752547001310045741864064482994858725517447466956518791269169932445648176733222571493149677633458466238303338202397024368594782876418757885729107101337003000942293335972927791914092128049015459762627910570552481588840517794181929052167695766087488155678601288183543542923073978101547857013284386127286201766539534449930019800629538936985500723286651317181135886613537472684585432548981137176605194616937916884425342594781263103889520479565943807153019112539648471126389007133628569101551453423329441284357220996286746119420951661002309740709965531900508158669911445442647882872642845017253320486483194578920399848938236367456182203750973485668474338872490493370316338265717607297788917989136673251906232471180372801739215723908227692280772924566627505383375006926077210593619421268920302567443565378008318306375933345023502569729065152853271943677560156660399164048825639676930792905029514886934137991251748566670747175149389790386533381395346848378086126737554383821108448976538368483182588363399173104558509056638462025014631311831087429077292622159430204291594740306101839816855066950261973761508571761199475875722129872053120607918649803615960923395941041186351688548839119185179061511562752936158490008721501922265117853150892510275280451512386037921846921215338292871369243215273327141574788295902601571954853164447945467502858402360002383447905203451080332820138038807089807348326201227952633606773669875783326254859449060219173688677862411205621098369850197290177157801120404586491539351157834995461006366357454485082418882790675313599505192062229760153765297973085881648731173082370598284894044874039320535929359764541655607954724778620299692329561389719894679422187273605123365595211331087787582288795975803204596084790245063851941743126163775104599211024868794963417068620929088930685252348056925998333775103901013166178123051145719327066291671254465121517468025481903583516889717075706778656188008220346836321018130262329960275994035799977740462449521145315883703579044832931500072461734173558055678321534543411700202585608091662941986374015145695722728369219632295111877625307534025947814482046574602884855000628069348113982760168555840795421620575435572915106415375929390228843561207926437055600623679865443824643739469724719459965557955058380348255978396827760847315302517889517186307227611036305093600742622617173630586132915440246954329046162586917746305785076749374879923291817501634840688134655343709975893536074051729094126976575932951568186247471276364688365517570183534172746626073065104511957628663499228486787805910851189856535554349587616640164475880286336297040462890970677362565843002353147494612339120686321466370878446992104275415694109122465685712047172411333784898167640969249816334211768571503116710400681753031921154156119580425706586931272762137106974722260296555246110537155545324997508432752001992143019105053629960070429632978051030666506387862681576587726837451289768507963663710593809112254288358391941211547737599813019216509521401333060709873137329265181692268450634439540567298120315463923249817937804691037934221694952291007930299492375072993250630509428139027930841344730614116433556147640931044259184813639305423693789765205264563476483182726333715121120306292338892864879492097378478618848682608046473195392008403983080088038690495574197562192939221108257663976813610444900247209483403267967688376213967440757138872928630798218493143438797780887379588968409461434159271317578365114578289355818599029235343888888465874521308381377794436361197628390368945957601203165022798579015453447473527069728514545998614229027372911314637820455162254475353567736227936485450357102086445412089842350389087702230398493802147348096874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Now another approach I have applied is through Divide and Concur via recursion
i.e. Fib(n) = fib(n-1) + Fib(n-2) and then further recursion for n-1 & n-2.....till 2 & 1. which is programmed as -
package com.company.dynamicProgramming;
import java.math.BigInteger;
public class FibonacciByBigDecimal {
public static void main(String ...args) {
int n = 100000;
BigInteger[] fibOfnS = new BigInteger[n + 1];
System.out.println("fibonacci of "+ n + " is : " + fibByDivCon(n, fibOfnS));
}
static BigInteger fibByDivCon(int n, BigInteger[] fibOfnS){
if(fibOfnS[n]!=null){
return fibOfnS[n];
}
if (n == 1 || n== 2){
fibOfnS[n] = BigInteger.ONE;
return BigInteger.ONE;
}
// creates 2 further entries in stack
BigInteger fibOfn = fibByDivCon(n-1, fibOfnS).add( fibByDivCon(n-2, fibOfnS)) ;
fibOfnS[n] = fibOfn;
return fibOfn;
}
}
When i ran the code for n = 100,000 the result is as below -
Exception in thread "main" java.lang.StackOverflowError
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
at com.company.dynamicProgramming.FibonacciByBigDecimal.fibByDivCon(FibonacciByBigDecimal.java:29)
Above you can see the StackOverflowError is created. Now the reason for this is too many recursion as -
// creates 2 further entries in stack
BigInteger fibOfn = fibByDivCon(n-1, fibOfnS).add( fibByDivCon(n-2, fibOfnS)) ;
So each entry in stack create 2 more entries and so on... which is represented as -
Eventually so many entries will be created that system is unable to handle in the stack and StackOverflowError thrown.
For Prevention :
For Above example perspective
Avoid using recursion approach or reduce/limit the recursion by again one level division like if n is too large then split the n so that system can handle with in its limit.
Use other approach, like the loop approach I have used in 1st code sample. (I am not at all intended to degrade Divide & Concur or Recursion as they are legendary approaches in many most famous algorithms.. my intention is to limit or stay away from recursion if I suspect stack overflow issues)
Considering this was tagged with "hacking", I suspect the "stack overflow" he's referring to is a call stack overflow, rather than a higher level stack overflow such as those referenced in most other answers here. It doesn't really apply to any managed or interpreted environments such as .NET, Java, Python, Perl, PHP, etc, which web apps are typically written in, so your only risk is the web server itself, which is probably written in C or C++.
Check out this thread:
https://stackoverflow.com/questions/7308/what-is-a-good-starting-point-for-learning-buffer-overflow
Stack overflow occurs when your program uses up the entire stack. The most common way this happens is when your program has a recursive function which calls itself forever. Every new call to the recursive function takes more stack until eventually your program uses up the entire stack.