When I create a new variable in a C++ program, eg a char:
char c = 'a';
how does C++ then have access to this variable in memory? I would imagine that it would need to store the memory location of the variable, but then that would require a pointer variable, and this pointer would again need to be accessed.
See the docs:
When a variable is declared, the memory needed to store its value is
assigned a specific location in memory (its memory address).
Generally, C++ programs do not actively decide the exact memory
addresses where its variables are stored. Fortunately, that task is
left to the environment where the program is run - generally, an
operating system that decides the particular memory locations on
runtime. However, it may be useful for a program to be able to obtain
the address of a variable during runtime in order to access data cells
that are at a certain position relative to it.
You can also refer this article on Variables and Memory
The Stack
The stack is where local variables and function parameters reside. It
is called a stack because it follows the last-in, first-out principle.
As data is added or pushed to the stack, it grows, and when data is
removed or popped it shrinks. In reality, memory addresses are not
physically moved around every time data is pushed or popped from the
stack, instead the stack pointer, which as the name implies points to
the memory address at the top of the stack, moves up and down.
Everything below this address is considered to be on the stack and
usable, whereas everything above it is off the stack, and invalid.
This is all accomplished automatically by the operating system, and as
a result it is sometimes also called automatic memory. On the
extremely rare occasions that one needs to be able to explicitly
invoke this type of memory, the C++ key word auto can be used.
Normally, one declares variables on the stack like this:
void func () {
int i; float x[100];
...
}
Variables that are declared on the stack are only valid within the
scope of their declaration. That means when the function func() listed
above returns, i and x will no longer be accessible or valid.
There is another limitation to variables that are placed on the stack:
the operating system only allocates a certain amount of space to the
stack. As each part of a program that is being executed comes into
scope, the operating system allocates the appropriate amount of memory
that is required to hold all the local variables on the stack. If this
is greater than the amount of memory that the OS has allowed for the
total size of the stack, then the program will crash. While the
maximum size of the stack can sometimes be changed by compile time
parameters, it is usually fairly small, and nowhere near the total
amount of RAM available on a machine.
Assuming this is a local variable, then this variable is allocated on the stack - i.e. in the RAM. The compiler keeps track of the variable offset on the stack. In the basic scenario, in case any computation is then performed with the variable, it is moved to one of the processor's registers and the CPU performs the computation. Afterwards the result is returned back to the RAM. Modern processors keep whole stack frames in the registers and have multiple levels of registers, so it can get quite complex.
Please note the "c" name is no more mentioned in the binary (unless you have debugging symbols). The binary only then works with the memory locations. E.g. it would look like this (simple addition):
a = b + c
take value of memory offset 1 and put it in the register 1
take value of memory offset 2 and put in in the register 2
sum registers 1 and 2 and store the result in register 3
copy the register 3 to memory location 3
The binary doesn't know "a", "b" or "c". The compiler just said "a is in memory 1, b is in memory 2, c is in memory 3". And the CPU just blindly executes the commands the compiler has generated.
C++ itself (or, the compiler) would have access to this variable in terms of the program structure, represented as a data structure. Perhaps you're asking how other parts in the program would have access to it at run time.
The answer is that it varies. It can be stored either in a register, on the stack, on the heap, or in the data/bss sections (global/static variables), depending on its context and the platform it was compiled for: If you needed to pass it around by reference (or pointer) to other functions, then it would likely be stored on the stack. If you only need it in the context of your function, it would probably be handled in a register. If it's a member variable of an object on the heap, then it's on the heap, and you reference it by an offset into the object. If it's a global/static variable, then its address is determined once the program is fully loaded into memory.
C++ eventually compiles down to machine language, and often runs within the context of an operating system, so you might want to brush up a bit on Assembly basics, or even some OS principles, to better understand what's going on under the hood.
Lets say our program starts with a stack address of 4000000
When, you call a function, depending how much stack you use, it will "allocate it" like this
Let's say we have 2 ints (8bytes)
int function()
{
int a = 0;
int b = 0;
}
then whats gonna happen in assembly is
MOV EBP,ESP //Here we store the original value of the stack address (4000000) in EBP, and we restore it at the end of the function back to 4000000
SUB ESP, 8 //here we "allocate" 8 bytes in the stack, which basically just decreases the ESP addr by 8
so our ESP address was changed from
4000000
to
3999992
that's how the program knows knows the stack addresss for the first int is "3999992" and the second int is from 3999996 to 4000000
Even tho this pretty much has nothing to do with the compiler, it's really important to know because when you know how stack is "allocated", you realize how cheap it is to do things like
char my_array[20000];
since all it's doing is just doing sub esp, 20000 which is a single assembly instruction
but if u actually use all those bytes like memset(my_array,20000) that's a different history.
how does C++ then have access to this variable in memory?
It doesn't!
Your computer does, and it is instructed on how to do that by loading the location of the variable in memory into a register. This is all handled by assembly language. I shan't go into the details here of how such languages work (you can look it up!) but this is rather the purpose of a C++ compiler: to turn an abstract, high-level set of "instructions" into actual technical instructions that a computer can understand and execute. You could sort of say that assembly programs contain a lot of pointers, though most of them are literals rather than "variables".
int A[10000000]; //This gives a segmentation fault
int *A = (int*)malloc(10000000*sizeof(int));//goes without any set fault.
Now my question is, just out of curiosity, that if ultimately we are able to allocate higher space for our data structures, say for example, BSTs and linked lists created using the pointers approach in C have no as such memory limit(unless the total size exceeds the size of RAM for our machine) and for example, in the second statement above of declaring a pointer type, why is that we can't have an array declared of higher size(until it reaches the memory limit!!)...Is this because the space allocated is contiguous in a static sized array?.But then from where do we get the guarantee that in the next 1000000 words in RAM no other piece of code would be running...??
PS: I may be wrong in some of the statements i made..please correct in that case.
Firstly, in a typical modern OS with virtual memory (Linux, Windows etc.) the amount of RAM makes no difference whatsoever. Your program is working with virtual memory, not with RAM. RAM is just a cache for virtual memory access. The absolute limiting factor for maximum array size is not RAM, it is the size of the available address space. Address space is the resource you have to worry about in OSes with virtual memory. In 32-bit OSes you have 4 gigabytes of address space, part of which is taken up for various household needs and the rest is available to you. In 64-bit OSes you theoretically have 16 exabytes of address space (less than that in practical implementations, since CPUs usually use less than 64 bits to represent the address), which can be perceived as practically unlimited.
Secondly, the amount of available address space in a typical C/C++ implementation depends on the memory type. There's static memory, there's automatic memory, there's dynamic memory. The address space limits for each memory type are pre-set in advance by the compiler. Which raises the question: where are you declaring your large array? Which memory type? Automatic? Static? You provided no information, but this is absolutely necessary. If you are attempting to declare it as a local variable (automatic memory), then no wonder it doesn't work, since automatic memory (aka "stack memory") has very limited address space assigned to it. Your array simply does not fit. Meanwhile, malloc allocates dynamic memory, which normally has the largest amount of address space available.
Thirdly, many compilers provide you with options that control the initial distribution of address space between different kinds of memory. You can request a much larger stack size for your program by manipulating such options. Quite possibly you can request a stack so large, than your local array will fit in it without any problems. But in practice, for obvious reasons, it makes very little sense to declare huge arrays as local variables.
Assuming local variables, this is because on modern implementations automatic variables will be allocated on the stack which is very limited in space. This link gives some of the common stack sizes:
platform default size
=====================================
SunOS/Solaris 8172K bytes
Linux 8172K bytes
Windows 1024K bytes
cygwin 2048K bytes
The linked article also notes that the stack size can be changed for example in Linux, one possible way from the shell before running your process would be:
ulimit -s 32768 # sets the stack size to 32M bytes
While malloc on modern implementations will come from the heap, which is only limited to the memory you have available to the process and in many cases you can even allocate more than is available due to overcommit.
I THINK you're missing the difference between total memory, and your programs memory space. Your program runs in an environment created by your operating system. It grants it a specific memory range to the program, and the program has to try to deal with that.
The catch: Your compiler can't 100% know the size of this range.
That means your compiler will successfully build, and it will REQUEST that much room in memory when the time comes to make the call to malloc (or move the stack pointer when the function is called). When the function is called (creating a stack frame) you'll get a segmentation fault, caused by the stack overflow. When the malloc is called, you won't get a segfault unless you try USING the memory. (If you look at the manpage for malloc() you'll see it returns NULL when there's not enough memory.)
To explain the two failures, your program is granted two memory spaces. The stack, and the heap. Memory allocated using malloc() is done using a system call, and is created on the heap of your program. This dynamically accepts or rejects the request and returns either the start address, or NULL, depending on a success or fail. The stack is used when you call a new function. Room for all the local variables is made on the stack, this is done by program instructions. Calling a function can't just FAIL, as that would break program flow completely. That causes the system to say "You're now overstepping" and segfault, stopping the execution.
In programming languages like C and C++, people often refer to static and dynamic memory allocation. I understand the concept but the phrase "All memory was allocated (reserved) during compile time" always confuses me.
Compilation, as I understand it, converts high level C/C++ code to machine language and outputs an executable file. How is memory "allocated" in a compiled file ? Isn't memory always allocated in the RAM with all the virtual memory management stuff ?
Isn't memory allocation by definition a runtime concept ?
If I make a 1KB statically allocated variable in my C/C++ code, will that increase the size of the executable by the same amount ?
This is one of the pages where the phrase is used under the heading "Static allocation".
Back To Basics: Memory allocation, a walk down the history
Memory allocated at compile-time means the compiler resolves at compile-time where certain things will be allocated inside the process memory map.
For example, consider a global array:
int array[100];
The compiler knows at compile-time the size of the array and the size of an int, so it knows the entire size of the array at compile-time. Also a global variable has static storage duration by default: it is allocated in the static memory area of the process memory space (.data/.bss section). Given that information, the compiler decides during compilation in what address of that static memory area the array will be.
Of course that memory addresses are virtual addresses. The program assumes that it has its own entire memory space (From 0x00000000 to 0xFFFFFFFF for example). That's why the compiler could do assumptions like "Okay, the array will be at address 0x00A33211". At runtime that addresses are translated to real/hardware addresses by the MMU and OS.
Value initialized static storage things are a bit different. For example:
int array[] = { 1 , 2 , 3 , 4 };
In our first example, the compiler only decided where the array will be allocated, storing that information in the executable.
In the case of value-initialized things, the compiler also injects the initial value of the array into the executable, and adds code which tells the program loader that after the array allocation at program start, the array should be filled with these values.
Here are two examples of the assembly generated by the compiler (GCC4.8.1 with x86 target):
C++ code:
int a[4];
int b[] = { 1 , 2 , 3 , 4 };
int main()
{}
Output assembly:
a:
.zero 16
b:
.long 1
.long 2
.long 3
.long 4
main:
pushq %rbp
movq %rsp, %rbp
movl $0, %eax
popq %rbp
ret
As you can see, the values are directly injected into the assembly. In the array a, the compiler generates a zero initialization of 16 bytes, because the Standard says that static stored things should be initialized to zero by default:
8.5.9 (Initializers) [Note]:
Every object of static storage duration is zero-initialized at
program startup before any other initial- ization takes place. In some
cases, additional initialization is done later.
I always suggest people to disassembly their code to see what the compiler really does with the C++ code. This applies from storage classes/duration (like this question) to advanced compiler optimizations. You could instruct your compiler to generate the assembly, but there are wonderful tools to do this on the Internet in a friendly manner. My favourite is GCC Explorer.
Memory allocated at compile time simply means there will be no further allocation at run time -- no calls to malloc, new, or other dynamic allocation methods. You'll have a fixed amount of memory usage even if you don't need all of that memory all of the time.
Isn't memory allocation by definition a runtime concept?
The memory is not in use prior to run time, but immediately prior to execution starting its allocation is handled by the system.
If I make a 1KB statically allocated variable in my C/C++ code, will that increase the size of the executable by the same amount?
Simply declaring the static will not increase the size of your executable more than a few bytes. Declaring it with an initial value that is non-zero will (in order to hold that initial value). Rather, the linker simply adds this 1KB amount to the memory requirement that the system's loader creates for you immediately prior to execution.
Memory allocated in compile time means that when you load the program, some part of the memory will be immediately allocated and the size and (relative) position of this allocation is determined at compile time.
char a[32];
char b;
char c;
Those 3 variables are "allocated at compile time", it means that the compiler calculates their size (which is fixed) at compile time. The variable a will be an offset in memory, let's say, pointing to address 0, b will point at address 33 and c at 34 (supposing no alignment optimization). So, allocating 1Kb of static data will not increase the size of your code, since it will just change an offset inside it. The actual space will be allocated at load time.
Real memory allocation always happens in run time, because the kernel needs to keep track of it and to update its internal data structures (how much memory is allocated for each process, pages and so on). The difference is that the compiler already knows the size of each data you are going to use and this is allocated as soon as your program is executed.
Remember also that we are talking about relative addresses. The real address where the variable will be located will be different. At load time the kernel will reserve some memory for the process, lets say at address x, and all the hard coded addresses contained in the executable file will be incremented by x bytes, so that variable a in the example will be at address x, b at address x+33 and so on.
Adding variables on the stack that take up N bytes doesn't (necessarily) increase the bin's size by N bytes. It will, in fact, add but a few bytes most of the time.
Let's start off with an example of how adding a 1000 chars to your code will increase the bin's size in a linear fashion.
If the 1k is a string, of a thousand chars, which is declared like so
const char *c_string = "Here goes a thousand chars...999";//implicit \0 at end
and you then were to vim your_compiled_bin, you'd actually be able to see that string in the bin somewhere. In that case, yes: the executable will be 1 k bigger, because it contains the string in full.
If, however you allocate an array of ints, chars or longs on the stack and assign it in a loop, something along these lines
int big_arr[1000];
for (int i=0;i<1000;++i) big_arr[i] = some_computation_func(i);
then, no: it won't increase the bin... by 1000*sizeof(int)
Allocation at compile time means what you've now come to understand it means (based on your comments): the compiled bin contains information the system requires to know how much memory what function/block will need when it gets executed, along with information on the stack size your application requires. That's what the system will allocate when it executes your bin, and your program becomes a process (well, the executing of your bin is the process that... well, you get what I'm saying).
Of course, I'm not painting the full picture here: The bin contains information about how big a stack the bin will actually be needing. Based on this information (among other things), the system will reserve a chunk of memory, called the stack, that the program gets sort of free reign over. Stack memory still is allocated by the system, when the process (the result of your bin being executed) is initiated. The process then manages the stack memory for you. When a function or loop (any type of block) is invoked/gets executed, the variables local to that block are pushed to the stack, and they are removed (the stack memory is "freed" so to speak) to be used by other functions/blocks. So declaring int some_array[100] will only add a few bytes of additional information to the bin, that tells the system that function X will be requiring 100*sizeof(int) + some book-keeping space extra.
On many platforms, all of the global or static allocations within each module will be consolidated by the compiler into three or fewer consolidated allocations (one for uninitialized data (often called "bss"), one for initialized writable data (often called "data"), and one for constant data ("const")), and all of the global or static allocations of each type within a program will be consolidated by the linker into one global for each type. For example, assuming int is four bytes, a module has the following as its only static allocations:
int a;
const int b[6] = {1,2,3,4,5,6};
char c[200];
const int d = 23;
int e[4] = {1,2,3,4};
int f;
it would tell the linker that it needed 208 bytes for bss, 16 bytes for "data", and 28 bytes for "const". Further, any reference to a variable would be replaced with an area selector and offset, so a, b, c, d, and e, would be replaced by bss+0, const+0, bss+4, const+24, data+0, or bss+204, respectively.
When a program is linked, all of the bss areas from all the modules are be concatenated together; likewise the data and const areas. For each module, the address of any bss-relative variables will be increased by the size of all preceding modules' bss areas (again, likewise with data and const). Thus, when the linker is done, any program will have one bss allocation, one data allocation, and one const allocation.
When a program is loaded, one of four things will generally happen depending upon the platform:
The executable will indicate how many bytes it needs for each kind of data and--for the initialized data area, where the initial contents may be found. It will also include a list of all the instructions which use a bss-, data-, or const- relative address. The operating system or loader will allocate the appropriate amount of space for each area and then add the starting address of that area to each instruction which needs it.
The operating system will allocate a chunk of memory to hold all three kinds of data, and give the application a pointer to that chunk of memory. Any code which uses static or global data will dereference it relative to that pointer (in many cases, the pointer will be stored in a register for the lifetime of an application).
The operating system will initially not allocate any memory to the application, except for what holds its binary code, but the first thing the application does will be to request a suitable allocation from the operating system, which it will forevermore keep in a register.
The operating system will initially not allocate space for the application, but the application will request a suitable allocation on startup (as above). The application will include a list of instructions with addresses that need to be updated to reflect where memory was allocated (as with the first style), but rather than having the application patched by the OS loader, the application will include enough code to patch itself.
All four approaches have advantages and disadvantages. In every case, however, the compiler will consolidate an arbitrary number of static variables into a fixed small number of memory requests, and the linker will consolidate all of those into a small number of consolidated allocations. Even though an application will have to receive a chunk of memory from the operating system or loader, it is the compiler and linker which are responsible for allocating individual pieces out of that big chunk to all the individual variables that need it.
The core of your question is this: "How is memory "allocated" in a compiled file? Isn't memory always allocated in the RAM with all the virtual memory management stuff? Isn't memory allocation by definition a runtime concept?"
I think the problem is that there are two different concepts involved in memory allocation. At its basic, memory allocation is the process by which we say "this item of data is stored in this specific chunk of memory". In a modern computer system, this involves a two step process:
Some system is used to decide the virtual address at which the item will be stored
The virtual address is mapped to a physical address
The latter process is purely run time, but the former can be done at compile time, if the data have a known size and a fixed number of them is required. Here's basically how it works:
The compiler sees a source file containing a line that looks a bit like this:
int c;
It produces output for the assembler that instructs it to reserve memory for the variable 'c'. This might look like this:
global _c
section .bss
_c: resb 4
When the assembler runs, it keeps a counter that tracks offsets of each item from the start of a memory 'segment' (or 'section'). This is like the parts of a very large 'struct' that contains everything in the entire file it doesn't have any actual memory allocated to it at this time, and could be anywhere. It notes in a table that _c has a particular offset (say 510 bytes from the start of the segment) and then increments its counter by 4, so the next such variable will be at (e.g.) 514 bytes. For any code that needs the address of _c, it just puts 510 in the output file, and adds a note that the output needs the address of the segment that contains _c adding to it later.
The linker takes all of the assembler's output files, and examines them. It determines an address for each segment so that they won't overlap, and adds the offsets necessary so that instructions still refer to the correct data items. In the case of uninitialized memory like that occupied by c (the assembler was told that the memory would be uninitialized by the fact that the compiler put it in the '.bss' segment, which is a name reserved for uninitialized memory), it includes a header field in its output that tells the operating system how much needs to be reserved. It may be relocated (and usually is) but is usually designed to be loaded more efficiently at one particular memory address, and the OS will try to load it at this address. At this point, we have a pretty good idea what the virtual address is that will be used by c.
The physical address will not actually be determined until the program is running. However, from the programmer's perspective the physical address is actually irrelevant—we'll never even find out what it is, because the OS doesn't usually bother telling anyone, it can change frequently (even while the program is running), and a main purpose of the OS is to abstract this away anyway.
An executable describes what space to allocate for static variables. This allocation is done by the system, when you run the executable. So your 1kB static variable won't increase the size of the executable with 1kB:
static char[1024];
Unless of course you specify an initializer:
static char[1024] = { 1, 2, 3, 4, ... };
So, in addition to 'machine language' (i.e. CPU instructions), an executable contains a description of the required memory layout.
Memory can be allocated in many ways:
in application heap (whole heap is allocated for your app by OS when the program starts)
in operating system heap (so you can grab more and more)
in garbage collector controlled heap (same as both above)
on stack (so you can get a stack overflow)
reserved in code/data segment of your binary (executable)
in remote place (file, network - and you receive a handle not a pointer to that memory)
Now your question is what is "memory allocated at compile time". Definitely it is just an incorrectly phrased saying, which is supposed to refer to either binary segment allocation or stack allocation, or in some cases even to a heap allocation, but in that case the allocation is hidden from programmer eyes by invisible constructor call. Or probably the person who said that just wanted to say that memory is not allocated on heap, but did not know about stack or segment allocations.(Or did not want to go into that kind of detail).
But in most cases person just wants to say that the amount of memory being allocated is known at compile time.
The binary size will only change when the memory is reserved in the code or data segment of your app.
You are right. Memory is actually allocated (paged) at load time, i.e. when the executable file is brought into (virtual) memory. Memory can also be initialized on that moment. The compiler just creates a memory map. [By the way, stack and heap spaces are also allocated at load time !]
I think you need to step back a bit. Memory allocated at compile time.... What can that mean? Can it mean that memory on chips that have not yet been manufactured, for computers that have not yet been designed, is somehow being reserved? No. No, time travel, no compilers that can manipulate the universe.
So, it must mean that the compiler generates instructions to allocate that memory somehow at runtime. But if you look at it in from the right angle, the compiler generates all instructions, so what can be the difference. The difference is that the compiler decides, and at runtime, your code can not change or modify its decisions. If it decided it needed 50 bytes at compile time, at runtime, you can't make it decide to allocate 60 -- that decision has already been made.
If you learn assembly programming, you will see that you have to carve out segments for the data, the stack, and code, etc. The data segment is where your strings and numbers live. The code segment is where your code lives. These segments are built into the executable program. Of course the stack size is important as well... you wouldn't want a stack overflow!
So if your data segment is 500 bytes, your program has a 500 byte area. If you change the data segment to 1500 bytes, the size of the program will be 1000 bytes larger. The data is assembled into the actual program.
This is what is going on when you compile higher level languages. The actual data area is allocated when it is compiled into an executable program, increasing the size of the program. The program can request memory on the fly, as well, and this is dynamic memory. You can request memory from the RAM and the CPU will give it to you to use, you can let go of it, and your garbage collector will release it back to the CPU. It can even be swapped to a hard disk, if necessary, by a good memory manager. These features are what high level languages provide you.
I would like to explain these concepts with the help of few diagrams.
This is true that memory cannot be allocated at compile time, for sure.
But, then what happens in fact at compile time.
Here comes the explanation.
Say, for example a program has four variables x,y,z and k.
Now, at compile time it simply makes a memory map, where the location of these variables with respect to each other is ascertained.
This diagram will illustrate it better.
Now imagine, no program is running in memory.
This I show by a big empty rectangle.
Next, the first instance of this program is executed.
You can visualize it as follows.
This is the time when actually memory is allocated.
When second instance of this program is running, the memory would look like as follows.
And the third ..
So on and so forth.
I hope this visualization explains this concept well.
There is very nice explanation given in the accepted answer. Just in case i will post the link which i have found useful.
https://www.tenouk.com/ModuleW.html
One among the many thing what a compiler does is that create and maintain a SYMTAB(Symbol Table under the section.symtab). This will be purely created and maintained by compilers using any Data Structure(List, Trees...etc) and not for the developers eyes. Any access request made by the developers this is where it will hit first.
Now about the Symbol Table,
We only need to know about the two columns Symbol Name and the Offset.
Symbol Name column will have the variable names and the offset column will have the offset value.
Lets see this with an example:
int a , b , c ;
Now we all know that the register Stack_Pointer(sp) points to the Top of the Stack Memory. Let that be sp = 1000.
Now the Symbol Name column will have three values in it a then b and then c. Reminding you all that variable a will be at the top of the stack memory.
So a's equivalent offset value will be 0.
(Compile Time Offset_Value)
Then b and its equivalent offset value will be 1. (Compile Time Offset_Value)
Then c and its equivalent offset value will be 2. (Compile Time Offset_Value)
Now calculating a's Physical address (or) Runtime Memory Address = (sp + offset_value of a)
= (1000 + 0) = 1000
Now calculating b's Physical address (or) Runtime Memory Address = (sp - offset_value of b)
= (1000 - 1) = 996
Now calculating c's Physical address (or) Runtime Memory Address = (sp - offset_value of c)
= (1000 - 2) = 992
Therefore at the time of the compilation we will only be having the offset values and only during the runtime the actual physical addresses are calculated.
Note:
Stack_Pointer value will be assigned only after the program is loaded. Pointer Arithmetic happens between the Stack_Pointer register and the variables offset to calculate the variables Physical Address.
"POINTERS AND POINTER ARITHMETIC, WAY OF THE PROGRAMMING WORLD"
Share what I learned about this question.
You can understand this issue in two steps:
First, the compilation step: the compiler generates the binary. In Linux system, binary is a file in ELF (Executable and Linkable Format) format. ELF file contains several sections, including .bss and .data
.data
Initialized data, with read/write access rights
.bss
Uninitialized data, with read/write access rights (=WA)
.data and .bss just map to the segments of process's memory layout, which contains static variables.
second, the loading step. When the binary file get executed, the ELF file will be loaded into process's memory. The loader can find static variables' information from ELF file.
Simply speaking, the compiler and the loader follow the same standard to communicate with each other, and the standard is ELF format.
As far as I know, the near pointers (or not far pointers) in C/C++ have address value which is very small than the actual address in RAM. So, if I keep on incrementing a pointer (say int type or any object type pointer) then at particular value it will be roll over. Now, my question is that: Is after rolling back the value pointed by it valid or not (assuming, I have a large size of data in the memory)?
I know this is a strange question to ask but I have a situation where I am continuously allocating and deallocating memory. I am finding that at particular point the binary crashes due to invalid address value like
0x20, 0x45 or 0x10101 etc.
I was wondering that the issue is due to roll over of the pointer value and since the address is getting rollover due to pointer therefore it is showing invalid address and crashing when it is being accessed.
I hope the situation I am referring is similar to the question is being asked. Even if they are different, I would like to know answers to both. I tried searching on "continuous incrementing pointers" but didn't find my answer.
EDIT: This is a new code compiled with G++ 4.1.2 20080704 (Red Hat 4.1.2-48) on Red Hat linux.
Actually the code is very large to share. But I can brief it in words:
There are 3 threads:
First thread: It creates allocates Alert class object and pushes it into the queue.
Second thread: It reads Alert from the queue, process it.
Third thread: It release the memory allocated to Alert objects after 20-30 minutes of processing.
I have already verified that the 3rd thread is not deallocating it before processed by 2nd thread.
But since the Alerts are generated on regular basis (i.e. around Thousands in a second) so I was suspecting the issue mentioned in the main question.
Points to note in my implementation:
I am using linux pipe queue to push it from one thread to other. For that I am pushing only address value of the object from sender side, and ensured to not delete the object there. Is this a possible way of corruption? Following is the code of this particular task:
Alert* l_alert = new Alert(ADD_ACTION,
l_vehicleType,
l_vehicleNo,
l_projPolyline,
l_speed,
l_slotId);
m_ResultHandler->SendToWorker(&l_alert);
Implementation of queue functions:
S32 SendToWorker(queueDataType *p_instPtr)
{
S32 ret_val=SUCCESS;
QueueObj.Lock();
ret_val = QueueObj.Signal();
QueueObj.push(*p_instPtr);
QueueObj.UnLock();
return ret_val;
}
S32 GetFromReceiver(queueDataType *p_instPtr)
{
QueueObj.Lock();
while(QueueObj.size() == 0)
QueueObj.Wait();
*p_instPtr = QueueObj.front();
QueueObj.pop();
QueueObj.UnLock();
return SUCCESS;
}
Receiver End:
m_alertQueue->GetFromReceiver(&l_alert)
What is the OS? Are you using virtual memory? The C standard says that a pointer is allowed to point to one address past the end of an array (but not dereferenced).
Pointing anywhere else is undefined behaviour.
The concept "near" and "far" pointers is a concept that mainly exists in compilers for x86 in 16-bit mode, where a "near" pointer is a 16-bit offset to a default segment of some sort, and a "far" pointer has a segment and offset value in the pointer itself. In 32- and 64-bit OS's, pointers are (generally) just an offset within a flat memory model (all segments are based at address zero).
A pointer can, according to the C standard point "to single object or an array of elements, and one past that". Anything else is undefined behaviour by the standard. One reason for this statement is to support segmented memory, where pointers may not be easy to compare between different segments (in particular not if the segments don't have a direct base-address, such as in OS/2 1.x, which used 16-bit protected mode, so code doesn't have easy access to the base-address of a segment. Since segments CAN overlap, it's not possible to tell if base address A + offset A is the same or different from base address B + offset B).
What actually happens if you have a pointer that doesn't fulfil this criteria is as stated "undefined". In an x86 environment, the real answer is that "it won't crash, and nothing bad will happen if you read the memory", but of course, if you try to write to memory that isn't "yours", then something bad could happen. Exactly what depends on what memory you are overwriting, and what the use of that memory is. It's impossible to say exactly what happens without knowing exactly what the memory is used for, and what value is written to it.
In a modern 32- or 64-bit OS, accessing memory that is "invalid" will definitely cause the program to crash, because modern OS's have memory protection that prevent "wild memory accesses".