In my hardware I want to address 8GB RAM memory but compilation fails with error: "Value out of range for 32-bit array element". Is there any workaround that allow use 64bit numbers in memory property? I can't find binding file for this property in Documentation directory of kernel tree... Can anyone point me right place?
memory#200000000 {
device_type = "memory";
/* 8G RAM */
reg = <0x000000000 0x200000000 0x000000000 0x200000000>;
};
Thanks
Related
I'm trying to understand the way Rust deals with memory and I've a little program that prints some memory addresses:
fn main() {
let a = &&&5;
let x = 1;
println!(" {:p}", &x);
println!(" {:p} \n {:p} \n {:p} \n {:p}", &&&a, &&a, &a, a);
}
This prints the following (varies for different runs):
0x235d0ff61c
0x235d0ff710
0x235d0ff728
0x235d0ff610
0x7ff793f4c310
This is actually a mix of both 40-bit and 48-bit addresses. Why this mix? Also, can somebody please tell me why the addresses (2, 3, 4) do not fall in locations separated by 8-bytes (since std::mem::size_of_val(&a) gives 8)? I'm running Windows 10 on an AMD x-64 processor (Phenom || X4) with 24GB RAM.
All the addresses do have the same size, Rust is just not printing trailing 0-digits.
The actual memory layout is an implementation detail of your OS, but the reason that a prints a location in a different memory area than all the other variables is, that a actually lives in your loaded binary, because it is a value that can already be calculated by the compiler. All the other variables are calculated at runtime and live on the stack.
See the compilation result on https://godbolt.org/z/kzSrDr:
.L__unnamed_4 contains the value 5; .L__unnamed_5, .L__unnamed_6 and .L__unnamed_1 are &5 &&5 and &&&5.
So .L__unnamed_1 is what on your system is at 0x7ff793f4c310. While 0x235d0ff??? is on your stack and calculated in the red and blue areas of the code.
This is actually a mix of both 40-bit and 48-bit addresses. Why this mix?
It's not really a mix, Rust just doesn't display leading zeroes. It's really about where the OS maps the various components of the program (data, bss, heap and stack) in the address space.
Also, can somebody please tell me why the addresses (2, 3, 4) do not fall in locations separated by 8-bytes (since std::mem::size_of_val(&a) gives 8)?
Because println! is a macro which expands to a bunch of stuff in the stackframe, so your values are not defined next to one another in the frame final code (https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=5b812bf11e51461285f51f95dd79236b). Though even if they were there'd be no guarantee the compiler wouldn't e.g. be reusing now-dead memory to save up on frame size.
I have my code compiled for certain ARM processor and have the binary. Now I want to know the exact size in bytes (address range) it occupies on my FLASh memory.
Coz, I have certain recovery mechanism at the last 1kB of flash and don't want that to be overwritten as it needs to be there permanently.
readelf of binary gives me the start addresses ( mapped to the code & data segments) & I couldn't really map this to what I want.
Pre-initialize flash memory with value'ab', load binary. Read flash memory until you encounter more than 2 'ab' values. This should give the address range in flash memory occupied by binary. ( THis is with the assumption that your binary might not have more than 2 'ab' as part of the binary)
If your compiler/linker is based on gnu toolchain (gcc/ld)
1/ At Compile Time
In your linker script adjust the section size to substract 1K.
You compiler throw error if your code not fit into your flash area.
Example :
MEMORY
{
FLASH (rx) : ORIGIN = 0x08001000, LENGTH = 128K-1K
RAM (xrw) : ORIGIN = 0x20000000, LENGTH = 16K
}
2/ At Run Time
You can set a symbol in your linker script to determine the end of your program (text segment). You can use this symbol to make a runtime test
Example :
.text :
{
. = ALIGN(4);
_etext = .; /* define a global symbols at end of code */
} >FLASH
3/ Manually
After compiling, use objcopy to convert your elf file to get the binary image that go into your flash. Check your datasheet to get your flash size and manually check if the file size fit into you flash minus 1K.
Exemple :
objcopy -O binary myfile.elf myfile.bin
what is the maximum amount of memory for a single process in UNIX and Linux and windows? how to calculate that? How much user address space and kernel address space for 4 GB of RAM?
How much user address space and kernel address space for 4 GB of RAM?
The address space of a process is divided into two parts,
User space: On standard 32 bit x86_64 architecture,the maximum addressable memory is 4GB, out of which addresses from 0x00000000 to 0xbfffffff = (3GB) meant for code, data segments. This region can be addressed when user process executing either in user or kernel mode.
Kernel space: Similarly, addresses from 0xc0000000 to 0xffffffff = (1GB) are meant for virtual address space of the kernel and can only addressed when the process executes in kernel mode.
This particular address space split on x86 is determined by the value of PAGE_OFFSET. Referring to Linux 3.11.1v page_32_types.h and page_64_types.h, page offset is defined as below,
#define __PAGE_OFFSET _AC(CONFIG_PAGE_OFFSET, UL)
Where Kconfig defines a default value of default 0xC0000000 also with other address split options available.
Similarly for 64 bit,
#define __PAGE_OFFSET _AC(0xffff880000000000, UL).
On 64 bit architecture 3G/1G split doesn't hold anymore due to huge address space. As per the source latest Linux version has given above offset as offset.
When I see my 64 bit x86_64 architecture, a 32 bit process can have entire 4GB of user address space and kernel will hold address range above 4GB. Interestingly on modern 64 bit x86_64 CPU's not all address lines are enabled(or the address bus is not large enough) to provide us 2^64 = 16 exabytes of virtual address space. Perhaps AMD64/x86 architectures has 48/42 lower bits enabled respectively resulting to 2^48 = 256TB / 2^42= 4TB of address space. Now this definitely improves performance with large amount of RAM, at the same time question arises how it is efficiently managed with the OS limitations.
In Linux there's a way to find out the limit of address space you can have.
Using the rlimit structure.
struct rlimit {
rlim_t cur; //current limit
rlim_t max; //ceiling for cur.
}
rlim_t is a unsigned long type.
and you can have something like:
#include <stdio.h>
#include <stdlib.h>
#include <sys/resource.h>
//Bytes To GigaBytes
static inline unsigned long btogb(unsigned long bytes) {
return bytes / (1024 * 1024 * 1024);
}
//Bytes To ExaBytes
static inline double btoeb(double bytes) {
return bytes / (1024.00 * 1024.00 * 1024.00 * 1024.00 * 1024.00 * 1024.00);
}
int main() {
printf("\n");
struct rlimit rlim_addr_space;
rlim_t addr_space;
/*
* Here we call to getrlimit(), with RLIMIT_AS (Address Space) and
* a pointer to our instance of rlimit struct.
*/
int retval = getrlimit(RLIMIT_AS, &rlim_addr_space);
// Get limit returns 0 if succeded, let's check that.
if(!retval) {
addr_space = rlim_addr_space.rlim_cur;
fprintf(stdout, "Current address_space: %lu Bytes, or %lu GB, or %f EB\n", addr_space, btogb(addr_space), btoeb((double)addr_space));
} else {
fprintf(stderr, "Coundn\'t get address space current limit.");
return 1;
}
return 0;
}
I ran this on my computer and... prrrrrrrrrrrrrrrrr tsk!
Output: Current address_space: 18446744073709551615 Bytes, or 17179869183 GB, or 16.000000 EB
I have 16 ExaBytes of max address space available on my Linux x86_64.
Here's getrlimit()'s definition
it also lists the other constants you can pass to getrlimits() and introduces getrlimit()s sister setrlimit(). There is when the max member of rlimit becomes really important, you should always check you don't exceed this value so the kernel don't punch your face, drink your coffee and steal your papers.
PD: please excuse my sorry excuse of a drum roll ^_^
On Linux systems, see man ulimit
(UPDATED)
It says:
The ulimit builtin is used to set the resource usage limits of the
shell and any processes spawned by it. If a new limit value is
omitted, the current value of the limit of the resource is printed.
ulimit -a prints out all current values with switch options, other switches, e.g. ulimit -n prints out no. of max. open files.
Unfortunatelly, "max memory size" tells "unlimited", which means that it is not limited by system administrator.
You can view the memory size by
cat /proc/meminfo
Which results something like:
MemTotal: 4048744 kB
MemFree: 465504 kB
Buffers: 316192 kB
Cached: 1306740 kB
SwapCached: 508 kB
Active: 1744884 kB
(...)
So, if ulimit says "unlimited", the MemFree is all yours. Almost.
Don't forget that malloc() (and new operator, which calls malloc()) is a STDLIB function, so if you call malloc(100) 10 times, there will be lot of "slack", follow link to learn why.
Have some issues with passing large amount of data (3 MB) from uboot to linux kernel 2.6.35.3 on imx50 ARM board. This data is required in kernel device driver probe function and then it should be released. First uboot load data from flash to RAM, then pass physical address for linux kernel using bootargs. In kernel I try to reserve certain amount of memory using reserve_resource() in arch/arm/kernel/setup.c file:
--- a/arch/arm/kernel/setup.c Tue Jul 17 11:22:39 2012 +0300
+++ b/arch/arm/kernel/setup.c Fri Jul 20 14:17:16 2012 +0300
struct resource my_mem_res = {
.name = "My_Region",
.start = 0x77c00000,
.end = 0x77ffffff,
.flags = IORESOURCE_MEM | IORESOURCE_BUSY,
};
## -477,6 +479,10 ##
kernel_code.end = virt_to_phys(_etext - 1);
kernel_data.start = virt_to_phys(_data);
kernel_data.end = virt_to_phys(_end - 1);
+ my_mem_res.start = mi->bank[i].start + mi->bank[i].size - 0x400000;
+ my_mem_res.end = mi->bank[i].start + mi->bank[i].size - 1;
for (i = 0; i < mi->nr_banks; i++) {
if (mi->bank[i].size == 0)
## -496,6 +502,8 ##
if (kernel_data.start >= res->start &&
kernel_data.end <= res->end)
request_resource(res, &kernel_data);
+
+ request_resource(res, &my_mem_res);
}
if (mdesc->video_start) {
By this I'm trying to tell kernel that this memory area it reserved and this data should not be modified by kernel.
70000000-77ffffff : System RAM
70027000-7056ffff : Kernel text
70588000-7062094f : Kernel data
77c00000-77ffffff : My_Region
In driver ioremap(0x77c00000, AREA_SIZE) is used to get kernel memory address. But when I dump content of memory, there is only zeros. If boot kernel with mem=120M (total 128MB RAM is avaliable), then my data is above kernel system ram region, then I get data I expect.
So, my questions:
Why I get zeros and how do I pass large amount of binary data from uboot to linux kernel?
You could use a custom ATAG to either pass the data block or to pass the address & length of the data. Note that the "A" in ATAG stands for ARM, so this solution is not portable to other architectures. An ATAG is preferable to a command-line bootarg IMO because you do not want the user to muck with physical memory addresses. Also the Linux kernel will process the ATAG list before the MMU (i.e. virtual memory) is enabled.
In U-Boot, look at lib_arm/armlinux.c or arch/arm/lib/bootm.c for routines that build the ARM tag list. Write your own routine for your new tag(s), and then invoke it in do_bootm_linux().
In the Linux kernel ATAGs are processed in arch/arm/kernel/setup.c, when virtual memory has not yet been enabled. If you just pass an address & length values from U-Boot, then the pointer & length can be assigned to global variables that are exported,
void *my_data;
unsigned int my_dlen;
EXPORT_SYMBOL(my_data);
EXPORT_SYMBOL(my_dlen);
and then the driver can retrieve it.
extern void *my_data;
extern unsigned int my_dlen;
request_mem_region(my_data, my_dlen, DRV_NAME);
md_map = ioremap(my_data, my_dlen);
I've used similar code to probe for SRAM in U-Boot, then pass the starting address & number of KBytes found to the kernel in a custom ATAG. A kernel driver obtains these values, and if they are nonzero and have sane values, creates a block device out of the SRAM. The major difference from your situation is that the SRAM is in a completely different physical address range from the SDRAM.
NOTE
An ATAG is built by U-Boot for the physical memory that the kernel can use, so this is really where you need to define and exclude your reserved RAM. It's probably too late to do that in the kernel.
I am trying to find the memory map of an array or some memory allocated from malloc() using mmap() but it is showing invalid argument.
#include<stdio.h>
#include<sys/mman.h>
#include<stdlib.h>
int main()
{
int *var1=NULL;
size_t size=0;
size = 1000*sizeof(int);
var1 = (int*)malloc(size);
int i=0;
for(i=0;i<999;i++)
{
var1[i] = 1;
}
printf("%p\n",var1);
void *addr=NULL;
addr = mmap((void *)var1, size, PROT_EXEC|PROT_READ|PROT_WRITE, MAP_SHARED | MAP_ANONYMOUS | MAP_FIXED, -1, 0); //to create memory map of var1
err(1,NULL); //to print error
return 0;
}
Error:
a.out: Invalid argument
Please help me.
Thank you in advance.
Proximate cause: mmap fails because you asked it do create a new memory mapping, you asked for the mapping to be placed at a specific address (var1's address), that address is already occupied (by the heap from which malloc got its memory), and you told the operating system it was not allowed to choose an alternate address in case var1 was not a suitable address (MAP_FIXED).
Analysis: What are you trying to do here? What does "find the memory map of an array" mean? Do you want to have your array of integers located in heap memory (returned by malloc()) or in an anonymous memory mapping created by mmap()? By the way, unless you fork() (create a child process) there is little functional difference: both are areas of memory that are private to your process. But they are not the same thing and you can't manipulate the heap with mmap() nor can you manage mapped memory with malloc().