I have a section of code mentioning the following.
Image$$PAGING_IMAGE_LOAD_END$$Base
What does this mean? An explanation or help of any kind would be great.
It's a special symbol generated by the ARM linker.
Image$$region_name$$Base Execution address of the region.
Image$$region_name$$Length Execution region length in bytes excluding ZI length.
Image$$region_name$$Limit Address of the byte beyond the end of the non-ZI part of the execution region.
[...etc...]
Related
For this C code:
foobar.c:
static int array[256];
int main() {
return 0;
}
the array is initialized to all 0's, by the C standard. However, when I compile
gcc -S foobar.c
this produces the assembly code foobar.s that I can inspect, and nowhere in foobar.s, is there any initialization of contents of the array.
Hence I reason, that the contents are not initialized, only when an element of the array is inspected, is it initialized, kind of like "copy-on-write" mechanism for fork.
Is my reasoning correct? If so, is this a documented feature, and if so where can I find that documentation?
There's kind of a lot of levels here. This answer addresses Linux in particular, but the same concepts are likely to apply on other systems, possibly with different names.
The compiler requires that the object be "zero initialized". In other words, when a memory read instruction is executed with an address in that range, the value that it reads must be zero. As you say, this is necessary to achieve the behavior dictated by the C standard.
The compiler accomplishes this by asking the assembler to fill the space with zeros, one way or another. It may use the .space or .zero directive which implicitly requests this. It will also place the object in a section with the special name .bss (the reasons for this name are historical). If you look further up in the assembly output, you should see a directive like .bss or .section .bss. The assembler and linker promises that this entire section will be (somehow) initialized to zero. This is documented:
The bss section is used for local common variable storage. You may allocate address space in the bss section, but you may not dictate data to load into it before your program executes. When your program starts running, all the contents of the bss section are zeroed bytes.
Okay, so now what do the assembler and linker do to make it happen? Well, an ELF executable file has a segment header, which specifies how and where code and data from the file should be mapped into the program's memory. (Please note that the use of the word "segment" here has nothing to do with the x86 memory segmentation model or segment registers, and is only vaguely related to the term "segmentation fault".) The size of the segment, and the amount of data to be mapped, are specified separately. If the size is greater, then all remaining bytes are to be initialized to zero. This is also documented in the above-linked man page:
PT_LOAD
The array element specifies a loadable segment,
described by p_filesz and p_memsz. The bytes
from the file are mapped to the beginning of the
memory segment. If the segment's memory size
p_memsz is larger than the file size p_filesz,
the "extra" bytes are defined to hold the value
0 and to follow the segment's initialized area.
So the linker ensures that the ELF executable contains such a segment, and that all objects in the .bss section are in this segment, but not within the part that is mapped to the file.
Once all this is done, then the observable behavior is guaranteed: as above, when an instruction attempts to read from this object before it has been written, the value it reads will be zero.
Now as to how that behavior is ensured at runtime: that is the job of the kernel. It could do it by pre-allocating actual physical memory for that range of virtual addresses, and filling it with zeros. Or by an "allocate on demand" method, like what you describe, by leaving those pages unmapped in the CPU's page tables. Then any access to those pages by the application will cause a page fault, which will be handled by the kernel, which will allocate zero-filled physical memory at that time, and then restart the faulting instruction. This is completely transparent to the application. It just sees that the read instruction got the value zero. If there was a page fault, then it just seems to the application like the read instruction took a long time to execute.
The kernel normally uses the "on demand" method, because it is more efficient in case not all of the "zero initialized" memory is actually used. But this is not going to be documented as guaranteed behavior; it is an implementation detail. An application programmer need not care, and in fact must not care, how it works under the hood. If the Linux kernel maintainers decide tomorrow to switch everything to the pre-allocate method, every application will work exactly as it did before, just maybe a little faster or slower.
Here says:
A typical memory representation of C program consists of following sections.
Text segment
Initialized data segment
Uninitialized data segment
Stack
Heap
But, how to get start address and end address of each section?
an easy way to get the beginning and ending address of each memory segment is to incorporate a linker command file and within that file place a label at each start and end of each memory segment.
For gcc (and ld) this link should tell you everything you need to know:
https://www.math.utah.edu/docs/info/ld_3.html
__eds__ WORD __ramspace[0x100] __attribute__((eds,address(0x8000ul),noload));
I want to understand the syntax above ( The program is for pic24 and in C ) especially __ramspace[0x100]. Can anybody help me?
It's a bit late, but maybe this can help someone else:
__eds__ means you want to put whatever follows into the extended data space. You do this when you want to use the data space beyond a certain address. You can find from which address the extended space begins for your MCU in the datasheet.
WORD means you will reserve whole words (and not for instance bytes). For a pic24 this means 16 bit chunks.
__ramspace[0x100] is a 1D array 256 in size. When you take a look at what is written in front of this, you can see you are declaring an array named __ramspace , size 256 words (so 256x 16 bit values) in extended data space (eds).
Now you must declare the offset e.g. the start address of the array (the physical address where __ramspace[0] - first array element - will be). This is what 0x8000 does.
Finally you instruct the compiler if the array should be initialised at boot-up (for instance filled with zeros). In your case there is a noload, meaning random data will be inside the array at boot-up, until you write your own values in it.
Hope this helps.
The __eds__ qualifier is described in the "MPLAB® C Compiler for PIC24 MCUs and dsPIC® DSCs User’s Guide" as:
In the attribute context the eds, for extended data space, attribute
indicates to the com- piler that the variable will may be allocated
anywhere within data memory. Variables with this attribute will likely
also need the eds type qualifier (see Chapter
6. “Additional C Pointer Types”) in order for the compiler to properly generate the correct access sequence. Note that the eds qualifier
and the eds attribute are closely related, but not identical. On some
devices, eds may need to be specified when allocating variables into
certain memory spaces such as space(ymemory) or space(dma) as this
memory may only exist in the extended data space.
__ramspace is not a special designator, it's just the identifier that was chosen.
__ramspace[0x100] is the only part of that line which is just pure C. :) It declares an array of 0x100 (256, in decimal) elements of type WORD. The name of the array is __ramspace.
See #Brian Cain's answer for details about eds.
The address(0x8000ul) argument to __attribute__() presumably makes the linker put the variable in question at location 0x8000.
This question already has an answer here:
Why do virtual memory addresses for linux binaries start at 0x8048000?
(1 answer)
Closed 8 years ago.
I am looking into to the memory layout of a given process. I notice that the starting memory location of each process is not 0. On this website, TEXT starts at 0x08048000. One reason can be to distinguish the address with the NULL pointer. I am just wondering if there is any another good reasons? Thanks.
The null pointer doesn't actually have to be 0. It's guaranteed in the C standard that when a 0 value is given in the context of a pointer it's treated as NULL by the compiler.
But the 0 that you use in your source code is just syntactic sugar that has no relation to the actual physical address the null-pointer value is "pointing" to.
For further details see:
Why is NULL/0 an illegal memory location for an object?
Why is address zero used for the null pointer?
An application on your operating system has its unique address space, which it sees as a continuous block of memory (the memory isn't physically continuous, it's just "the impression" the operating system gives to every program).
For the most part, each process's virtual memory space is laid out in a similar and predictable manner (this is the memory layout in a Linux process, 32-bit mode):
(image from Anatomy of a Program in Memory)
Look at the text segment (the default .text base on x86 is 0x08048000, chosen by the default linker script for static binding).
Why the magical 0x08048000? Likely because Linux borrowed that address from the System V i386 ABI.
... and why then did System V use 0x08048000?
The value was chosen to accommodate the stack below the .text section,
growing downward. The 0x48000 bytes could be mapped by the same page
table already required by the .text section (thus saving a page table
in most cases), while the remaining 0x08000000 would allow more room
for stack-hungry applications.
Is there anything below 0x08048000? There could be nothing (it's only 128M), but you can pretty much map anything you desire there, using the mmap() system call.
See also:
What's the memory before 0x08048000 used for in 32 bit machine?
Reorganizing the address space
mmap
I think this sums it up:
Each process has its own set of page tables, but there is a catch. Once virtual addresses are enabled, they apply to all software running in the machine, including the kernel itself. Thus a portion of the virtual address space must be reserved to the kernel.
So while the process gets it's own address space. Without allocating a block to the kernel, it would not be able to address kernel code and data.
This is always the first block of memory it appears and so includes address 0. The user mode space starts beyond this, and so that is where both the stack and heap reside.
Distinguishing from NULL pointer
Even if the user mode space started at address 0, there would not be any data allocated to the address 0 as that will be in the stack or the heap which themselves do not start at the beginning of the user area. Therefore NULL (with the value of 0) could be used still and is not a reason for this layout.
However one benefit related to the NULL and the first block being kernel memory is any attempt to read/write to NULL throws a Segmentation Fault.
A loader loads a binary in segments into memory: text (constants), data, code. There is no need to start from 0, and as C is has the problem from bugs accessing around null, like in a[i] that is even dangerous. This allows (on some processors) to intercept segmentation faults.
It would be the C runtime introducing a linear address space from 0. That might be imaginable where C is the operating system's implementation language. But serves no purpose; to have the heap start from 0. The memory model is one of segments. A code segment might be protected against modification by some processors.
And in segments allocation happens in C runtime managed memory blocks.
I might add, that physical 0 and upwards is often used by the operating system itself.
Im very new to embedded programming started yesterday actually and Ive noticed something I think is strange. I have a very simple program doing nothing but return 0.
int main() {
return 0;
}
When I run this in IAR Embedded Workbench I have a memory view showing me the programs memory. Ive noticed that in the memory there is some memory but then it is a big block of empty space and then there is memory again (I suck at explaining :P so here is an image of the memory)
Please help me understand this a little more than I do now. I dont really know what to search for because Im so new to this.
The first two lines are the 8 interrupt vectors, expressed as 32-bit instructions with the highest byte last. That is, read them in groups of 4 bytes, with the highest byte last, and then convert to an instruction via the usual method. The first few vectors, including the reset at memory location 0, turn out to be LDR instructions, which load an immediate address into the PC register. This causes the processor to jump to that address. (The reset vector is also the first instruction to run when the device is switched on.)
You can see the structure of an LDR instruction here, or at many other places via an internet search. If we write the reset vector 18 f0 95 e5 as e5 95 f0 18, then we see that the PC register is loaded with the address located at an offset of 0x20.
So the next two lines are memory locations referred to by instructions in the first two lines. The reset vector sends the PC to 0x00000080, which is where the C runtime of your program starts. (The other vectors send the PC to 0x00000170 near the end of your program. What this instruction is is left to the reader.)
Typically, the C runtime is code added to the front of your program that loads the global variables into RAM from flash, and sets the uninitialized RAM to 0. Your program starts after that.
Your original question was: why have such a big gap of unused flash? The answer is that flash memory is not really at a premium, so we can waste a little, and that having extra space there allows for forward-compatibility. If we need to increase the vector table size, then we don't need to move the code around. In fact, this interrupt model has been changed in the new ARM Cortex processors anyway.
Physical (not virtual) memory addresses map to physical circuits. The lowest addresses often map to registers, not RAM arrays. In the interest of consistency, a given address usually maps to the same functionality on different processors of the same family, and missing functionality appears as a small hole in the address mapping.
Furthermore, RAM is assigned to a contiguous address range, after all the I/O registers and housekeeping functions. This produces a big hole between all the registers and the RAM.
Alternately, as #Martin suggests, it may represent uninitialized and read-only Flash memory as -- bytes. Unlike truly unassigned addresses, access to this is unlikely to produce an exception, and you might even be able to make them "reappear" using appropriate Flash controller commands.
On a modern desktop-class machine, virtual memory hides all this from you, and even parts of the physical address map may be configurable. Many embedded-class processors allow configuration to the extent of specifying the location of the interrupt vector table.
UncleO is right but here is some additional information.
The project's linker command file (*.icf for IAR EW) determines where sections are located in memory. (Look under Project->Options->Linker->Config to identify your linker configuration file.) If you view the linker command file with a text editor you may be able to identify where it locates a section named .intvec (or similar) at address 0x00000000. And then it may locate another section (maybe .text) at address 0x00000080.
You can also see these memory sections identified in the .map file, along with their locations. (Ensure "Generate linker map file" is checked under Project->Options->Linker->List.) The map file is an output from the build, however, and it's the linker command file that determines the locations.
So that space in memory is there because the linker command file instructed it to be that way. I'm not sure whether that space is necessary but it's certainly not a problem. You might be able to experiment with the linker command file and move that second section around. But the exception table (a.k.a. interrupt vector table) must be located at 0x00000000. And you'll want to ensure that the reset vector points to the new location of the startup code if you move it.