It's the same thing from a conceptual point of view: it works on the principle of "the last come, the first left" (last in, first out).It is clear that the code also uses a structure/logic to track methods, but this is not a standard class library. They work at different levels.

At least in this sentenceprintf("Введите количество элементов\n"); scanf("%d",&SIZE); You use a name. SIZEwhich you don't have.And the most important thing is, you never have a glass, because it's in this function.void stack_creation(Item*STACK_AR,int STACK_SIZE) { STACK_AR=(Item*)malloc(STACK_SIZE*sizeof(Item)); } Parameter STACK_AR - it's not a global variable, as you think, all over the vilability, which is a local variable function that contains copy global variable with the same name STACK_ARand to be removed after separation. Therefore, changing this local variable inside the function does not affect the global variable.You should have identified this function in a subsequent way.void stack_creation(Item **STACK_AR,int STACK_SIZE) { *STACK_AR=(Item*)malloc(STACK_SIZE*sizeof(Item)); } I mean, you're making a point.This function should be triggered accordingly.stack_creation( &STACK_AR, STACK_SIZE ); EDIT: I see that you've downloaded the code and replaced it. SIZE ha STACK_SIZE♪ So my first comment can be ignored.

CurrentStatus := DataForm.Create; CurrentStatus.setForm('fRoot'); CurrentStatus.setOperation(oRoot); TopForm.Push(CurrentStatus); CurrentStatus := DataForm.Create; CurrentStatus.setForm('fViewPatientStory'); CurrentStatus.setOperation(View); TopForm.Push(CurrentStatus); CurrentStatus := DataForm.Create; CurrentStatus.setForm('fViewPatient'); CurrentStatus.setOperation(View); ShowMessage(TopForm.Top().getForm()); TopForm.Pop; ShowMessage(TopForm.Top().getForm()); And don't forget to release the memory of the three CurrentStatus if TStack doesn't do it.

Who's gonna sign zero at the end? You can write that.while(st[i]!='\0'&&i<100) { s.mas[s.top][i]=st[i]; i++; } s.mas[s.top][i] = '\0'; a can be found about the function strcpy and the joy of life (it replicates the hire and often a few times faster than the normal cycle).

If simple terms are to be clarified when echo/print is produced, the data are to be found (e.g. browser). In order to prevent multiple echos from racing the bayta, there is a weekend buffer where the data are accumulated to some volume and when overcrowded, the data are released. Once again, data are accumulated. flush forcibly ejects the contents of the buffer into the drain.For example, you have a long task where you have to keep sending the results to the browser and you use echo information. But they're digging up in the buffer and the browser won't get anything until the buffer gets over. There's also a way to turn off the buffer, and there's no cli in it.ob_* enables the initialization of its encumbered buffers by sending them to each other. And when it's cleaned, the data buffer will be down to the default. You can't make flush at all, but just take the results from the buffer and destroy it.ob_start opens a new buffer and all subsequent print/echo will enter it. At any time, you can open a new buffer, and you can close and drop it on the buffer open earlier. I mean, it looks like a elephant pie (scream) of buffers, where the data are written in the top buffer and only when all the buffers are dropped to the bottom of the default, until the data are clear.It's convenient to use it in a self-writing template. For a segment of the template, we create a buffer and then we take a ready-match and a buffer. Or when we call a side code that beats echo and creates trash, we call him in ob_* and all that trash is sitting there.I mean flush dumps. systems Buffer to the conclusion, and ob_flush drops last open Buffer through ob_start below buffer (ob_* or system)

Don't read Soviet newspapers before lunch.The default steak for the 32-bit code is 1 Mb, for 64-bit: 4 Mb.With the help of your designer, the size of the stack can be applied by anyone, but with some limitations: The minimum size (256 Kb) will not be taken into account and the minimum number of memory will be allocatedThere should be a continuous free memory block of the size indicated.The stack size is rounded up by page size (usually 64 CB)The code has to work in full trust if you want to ask more default.If you don't enter the processor and the processor architecture for you, you don't need the above information. https://blogs.msdn.microsoft.com/oldnewthing/20070301-00/?p=27803 ) https://stackoverflow.com/a/2902956/1985167 :class Program { private struct MEMORY_BASIC_INFORMATION { public uint BaseAddress; public uint AllocationBase; public uint AllocationProtect; public uint RegionSize; public uint State; public uint Protect; public uint Type; } private const uint STACK_RESERVED_SPACE = 4096 * 16; private const int MEGABYTE = 1024 * 1024; [DllImport("kernel32.dll")] private static extern int VirtualQuery( IntPtr lpAddress, ref MEMORY_BASIC_INFORMATION lpBuffer, int dwLength); static void Main(string[] args) { Console.WriteLine("Main thread"); Console.WriteLine(EstimatedRemainingStackSizeInMegabytes()); Console.WriteLine("New thread"); var thread = new Thread( () =&gt; Console.WriteLine(EstimatedRemainingStackSizeInMegabytes()), 16 * MEGABYTE); thread.Start(); thread.Join(); Console.ReadLine(); } private unsafe static float EstimatedRemainingStackSizeInMegabytes() { MEMORY_BASIC_INFORMATION stackInfo = new MEMORY_BASIC_INFORMATION(); IntPtr currentAddr = new IntPtr((uint)&amp;stackInfo - 4096); VirtualQuery(currentAddr, ref stackInfo, sizeof(MEMORY_BASIC_INFORMATION)); return (float)(currentAddr.ToInt64() - stackInfo.AllocationBase - STACK_RESERVED_SPACE) / MEGABYTE; } } Result:Main thread0.9277534New thread15.93015

For s+ x86 32-bit responseCONTEXT ctx; ctx.Flags = CONTEXT_CONTROL; if (GetThreadContext(GetCurrentThread(),&ctx)) { int * stack = (int*) ctx.Esp; // получили массив стека, в котором можно увидеть значения локальных переменных // тут можно смотреть стек printf("%x", stack[0]); } For some versions with++, an index can be obtainedint * stack = (int*)_ESP; // borland c++ int * stack; // для тех версий что поддерживают asm, этот вариант так же не все поддерживают asm { mov eax; esp; mov dword ptr stack, eax; } Many environments maintain the dumping of memory and cpu window - there are visual examinations of glasses.And the post office, the barbaric way. The glass is. It may be possible to use it and obtain an example of the steak address by taking the local variable (not equal to esp, but it can already be analysed):int a = 1; int b = 2; int * stack = (int*)&a; And watch stack[0]; stack[1] and stack[-1];On the organization, the stench liked: https://rsdn.ru/article/cpp/ObjectsAndPointers.xml with pictures.

isNumber()You don't need to invent bikes: http://www.opennet.ru/man.shtml?topic=isdigit&category=3 ♪If the symbol is the numberMaybe not a number, but a number?I don't have the same meaning, but his number is from table ASCII.In the glass symbol - exactly what you do with this code: A[top] = x; If the meaning is to be given, one figure is sufficient:A[top] = x - '0'; Which is like the code:A[top] = x - 48; // 48 - код символа '0' If there's a symbol in the x, say, "5', it'll be:A[top] = '5' - 48; // или A[top] = 53 - 48; // 53 - код символа '5' // оно же: A[top] = 5;

How to store in the glass, there are two stacks in this case. One for calculations, the other keeps the Polish record itself (although it's not the glass, it's just the turn).The calculation glass is best made int or double(float). The recording of the calculation itself is better made in the normal vector of the line (although it is possible in the glass as convenient). And keep it normal. I mean, I'd keep that. {"0", "a", "-', "0", "b", "-", "-"}♪How to do it. The code in the cycle reads the vector element and looks at it. If this number (first symbol is the figure) is placed in the computing glass if it is variable (a or b in your case, i.e., the first symbol is a..z A.Z.) - takes their meaning (e.g. the user asks) and wears it in the glass. If the operation in turn relies on two numbers, performs the operation and brings the result in the calculations. At the end of the job, the top of the line is just sealed.In fact, the code, I think you can write on your own.

Several technical councils:hand over complex objects to the reference (e.g., SumLongNumber(deque<int>& da, deque<int>& db)Because it's much faster (this time, everything's irreconcilable to the value, which has to slow down the process appreciably),operator+ It would be possible to make it more rapid if the avoidance of the creation of an intermediate facility (the same is the meaning of the operator),Ten is a slow operation. It is better to keep long whole numbers in structures with a good reason (twice level) and to translate into a decimal place at the last point (for the display),Don't use it. system("pause");because it adds dependence on PPs and it's dangerous at all (why does the system do when calling "pause").

N ROLL DROP where N - the cell number from the top of the line from 0.For example, the phrase 4 3 2 1 0 3 ROLL DROP left on the glass. 4 2 1 0♪Option https://forth-standard.org/standard/core/ROLL fixed on sheet starting with http://forth.sourceforge.net/standard/fst83/ ♪

There's a thing in the assembler that's got the notions of a near and far transition. The call is not different. The length of the transfer(s) is when the address is fully indicated (in this case, and the segment is moved). And the nearest (the short) transition is a one-by-side transition (up to 128 rearwards or 127 byte forward).ret It's just-- it's removing the address that kept the call there (or the user planted his hands, or the operating system built it all right) and making a transition to that address. Team http://www.mathemainzel.info/files/x86asmref.html#ret It's too close and far. And naturally, it has to be coupled to call.What is added in the glass? In the case of a major challenge, the contents of the registers cs and ip are added (i.e. 4 bayta). In the case of short (high) ip only (two baytes).And the third question. Yes, the registers are 16. And that's why there's a clean water called modullar arithmetic. Only the module didn't get much. It is 10000h. and in this case FFFEh + 0002h = 10000h x 0000h.Here. http://www.codenet.ru/progr/dos/int_0025.php - it's a DOS interruption. And that's how it turns out to be int 20 by a given shift, it's just a dos job.

Let's see the change in the stitch.Let esp == sync, corrected by elderman == Position of the post-objection position (the current esp is indicated by the arrow):-> X: YYYYYYYY ; в вершине стека что-то лежит, не важно что. After the team push DWORD PTR 2 esp decreases by 4 to the new esp address, recording 2:-> X-4: 2 X-0: YYYYYYYY After the team push DWORD PTR 3 esp again decreases by 4 at esp address 3:-> X-8: 3 X-4: 2 X-0: YYYYYYYY After the call. call AddDigs esp decreases by 4 to esp the return address:-> X-0Ch: RETADDR X-08h: 3 X-04h: 2 X-00h: YYYYYYYY When you leave the command procedure ret 8 The return address is removed from the stack, esp is increased by 4. The esp is then increased by the operand ret value (by 8): X-0Ch: RETADDR X-08h: 3 X-04h: 2 -> X-00h: YYYYYYYY Thus, esp-4 is located 2, esp-8 is 3 at esp-0Ch (esp-12), the return address.

That's it. Got it. You can help. setrlimit (RLIMIT_STACK, struct rlimit ...)♪

For com-fayla (tini memorial model), glass will in any case start from the end of the segment and increase to smaller addresses. If the size of the code+data of your program begins to approach 64k, we need to start thinking about moving to another model of memory (e.g. small model, expansion of the programme will be exe), where for the code, data, the grid can be distinguished by its own segments.

You're in method. full To the field. stkAnd this field is not initialized. You probably wanted to initial it in the designer:public Stek(int size) { char[] stk = new char[size]; t = 0; Console.WriteLine("создание объект"); } but in fact, instead of initializing the field stk You're creating a new, completely irrelevant set of instructions. char[] stk = new char[size]; If you want to initiate the class field, instead of char[] stk = new char[size] write stk = new char[size];Besides this mistake, you have a lot of problems in the code.

This "Thread.getStackTrace()" line is a method that returns a set of trace elements that are present in the current flow. The mass of zero length will return if the flow has not been started or has been completed.The top will be the last of the induced methods, the most recent element of the mass will be the recently launched method inside the flow.In case of error, usually caused getStackTrace()to catch the bug faster and track the sequence of intra-flow methods. Yeah, to the background, he needs to be looked up to the bottom to determine the bug or whatever. :Your question is why it's so thoughtful that this method would remove the sequence of other methods.

Glass is the data structure with access to LIFO elements (Last In First Out - The latter came - the first out). The data are added to the beginning (the end of which is convenient), and they are extracted. Only two functions and an index on the top are sufficient for this structure:push(item); //добавляет данные в стек pop(); //извлекает последний элемент The output is the data structure with access to FIFO elements (First In First Out - First Come - First come - First out). Data are added at the end and extracted from the beginning. Two indicators, one at the beginning, one at the turn, and one at the end of the data will be required for the rapid addition and extraction of the data. Functions are used to handle turns:enqueue(item); //добавляет новый элемент в очередь dequeue(); //извлекает элемент из очереди You need to see, for example, the C++ indicators. http://translated.by/you/a-beginner-s-guide-to-pointers/into-ru/ (one of the first in search).Then you need to set up the data element of your stack. For example:struct data { int p; int c; } After that, we'll have a grid. For example:struct element { data element_data; element *next; //указатель на следующий элемент } The indicator for the next element is necessary to ensure that our data in the glass(s) have some consistency and are somehow linked.Then write the necessary functions, start the indicators (and end, if necessary). For example, push()://somewhere in the code element *top; //инициализируется в NULL //... void push(data newData) { //создаем нашу структуру element и указатель на нее element *newElement = new element(); //element_data = newData element->element_data = newData; //next = top - здесь мы и задаем связку между двумя элементами стека element->next = top; //top = указателю на нашу структуру top = newElement; } Everything else is similar. I haven't coded for C++ a long time, so I'm sorry for all the guru in advance if I'm screwed.

Oh, why so many codes? If you think the problem is combinationYou're gonna have to use the example. combination plus hardcoded test data reproducing the problem.Code like printf("--------------------------------------------------------------\n"); definitely not required to understand the problem.Is it not hard for you to write a lesbian?By the way: http://www.sscce.org/

Usually, when they write a typewriter like that, they make a pattern.typedef enum { OPERATION, OPERAND, BRACE, ... // другие типы токенов, которые поддерживает ваш парсер } token_type; typedef struct { token_type type; union { char c; int i; ... // другие типы данных, которые может принимать токен } value; } token; The glass is not stored charYou, a tokenYou. When you separate the token, you put it in this structure and you put it right. token_type♪ It will then be possible to determine which type of token has been removed and produced by its type of action.