What is the role of ZTEXTADDR in Linux kernel ?
From lxr.linux.no, it's an address in RAM that holds address of zImage as sequence below?
A.
uImage (DataFlash/NAND) ---load_to_RAM--->
uImage (# boot_addr) ---decompress_uImage-->
zImage (# ZTEXTADDR) --- decompress_zImage--->
uncompressed image (# ZRELADDR).
or just:
B.
uImage (DataFlash/NAND) ---load_to_RAM--->
uImage (# boot_addr) ---decompress_uImage-->
uncompressed image (# ZRELADDR)
no using ZTEXTADDR in new kernel version for booting process ?
The Linux ARM decompression boot loader is capable of relocating itself when running from RAM. The relocation portion is PC-relative and so it can be loaded at any address. However, if your main image starts from FLASH/ROM, the code can not be relocated; while moving an image in RAM is a simple memmove(), it is much more involved for NOR flash and can be impossible for ROM.
In this case, a compressed boot linker script is used with the ZTEXTADDR as the location of the decompression code. In your diagram, you have a u-boot, which will load the uImage. There is no reason to execute this directly from Flash/ROM. u-boot can copy the image to RAM and there is no need for the ZTEXTADDR value and it should be left as zero.
If your image boots directly from Flash/ROM, without a boot loader then ZTEXTADDR is useful,
zImage (in flash) --> decompress vmlinux.bin to RAM --> run kernel
The zImage may need to be annotated with some chip setup for this to work and would need ATAGS or device trees linked. For this reason, there are many machine variants in boot/compressed; This is not maintainable and these type of files are discouraged. Typically another bootloader loads the image to RAM and the zImage can move itself to whatever destination it needs; I think that is your situation and you should set ZTEXTADDR to zero and forget about it.
Related
I am writing a bootloader for my Arm board (32-bit i.MX6) and want to boot the Linux kernel using a device tree and an initrd file located at a static location in memory.
I looked to U-Boot as a reference and I see I can use the bootm command to provide a kernel, device tree and ramdisk:
Boot application image from memory:
bootm [addr [arg ...]] - boot application image stored in memory passing arguments 'arg
...'; when booting a Linux kernel,‘arg' can be the address of an initrd image
I know there are two ways to pass information from a bootloader to the kernel, the legacy ATAGS method and the modern device tree method.
With ATAGS this can be achieved with the ATAG_INITRD2 tag which describes the address that a ramdisk is stored in memory. However, with a flattened device tree no ATAGS are passed to the kernel (which is shown in the boot log with the "No ATAGs?" message). I do not see any method to specify a ramdisk when using a device tree.
If I look at the documentation for booting the Linux kernel on Arm I see the following interface is specified:
- CPU register settings
r0 = 0,
r1 = machine type number discovered in (3) above.
r2 = physical address of tagged list in system RAM, or
physical address of device tree block (dtb) in system RAM
In fact, the same document states that an initramfs must be configured prior to booting, yet includes no details on how to do so when booting with a device tree.
Is there some alternative way to achieve this? Is the required information appended to the device tree automatically by U-Boot, or is there an alternative way to notify the kernel of the ramdisk location?
It will be patched in the dtree by the bootloader; e.g.
/ {
chosen {
linux,initrd-start = <0x....>;
linux,initrd-end = <0x....>;
};
I'm creating a bootable Linux system on a PicoZed board (ARM CortexA9 core), and I've run into a "limitation", which I don't think really is a limitation (I get the feeling it's another problem masquerading as a limitation).
I boot by starting the system in JTAG boot mode; after powering on the board I use the xmd debugger to place u-boot into the system's RAM and then I run it.
Next I place the kernel (uImage), the gzip'd initramfs image and the device tree into memory. Finally I tell u-boot to boot the system using bootm, and using three arguments to point out the memory locations of all the images.
All of this works, and I manage to boot up Linux + userland. However, I need to grow the initramfs, and this is where I run into problems.
The working image is 16MiB exactly. I tried to make it 24MiB (completely regenerated from scratch each time I try to boot), but just after the kernel has loaded and it tries to find init the kernel reports file system faults and fails. There shouldn't be any overlaps, but just in case I tried moving things around a little, but the same problem occurred.
After searching for some tips, I saw someone on a forum say that the image needs to be placed at a 16MiB alignment (which I don't think is true, but I tried it none the less, but it didn't get a working system). Another post claimed that the images must be aligned with their sizes (which I again don't think is true, but I tried that as well, but with no change). Yet another post claimed that this happens if the initramfs image crosses the __init end boundary, and that placing the initramfs image firmly inside will allow the memory to be reclaimed after the image has been uncompressed by the kernel, and placing it beyond the __init section will work but then that memory is forever lost after boot. I know way too little about Linux in order to know if any of this is in any way true/accurate, and I have no idea where "__init"'s - if such a thing exists - end-boundary is, but if the issue is that I was crossing it I tried moving the initramfs image way beyond anywhere were I was previously using it, but that didn't change anything.
I also tried the original location (which works with a 16MiB image) and create a 16MiB + 1K sized image, but this didn't work either. (Checking that it didn't overlay any of the over images, obviously).
This originally led me to think there's a 16MiB initramfs size limit lurking somewhere. But on searching for it, it got me thinking that doesn't make sense -- as far as I can gather the bootm command in u-boot shoud set up the tag list for the system (which includes the location and size of the initramfs), and I haven't come across any note about a 16MiB limit in relation to the tagged lists for the initramfs so far.
I have found a page which claims that the initramfs size is in practical terms limited to roughly half the size of physical ram in the system, and the PicoZed board has 1G RAM, so we're in orders of magnitude away from what "should" be a limitation.
To clarify: The 16MiB image is the size of the raw image; compressed it's just under 6MiB. However, if I populate it so it's not compressed as tightly it doesn't make any difference -- the problem doesn't appear to be related to the size of the compressed image, only the uncompressed image.
Main question:
Where is this apparent 16MiB initramfs size limit coming from?
Side-question:
Is there such a thing as an "kernel __init section" which is reclaimed by the kernel after it has uncompressed/loaded images? If so, how do I see/configure the location/size of it?
Size constraints of initramfs on ARM?
I have not encountered a 16 MiB size constraint as you allege.
At one time I thought there was a size limit too, but that turned out to be a memory footprint issue during boot. Once that was sorted out I've been using large initramfs of 30MiB (e.g. with glibc, gtstreamer and qt5 libraries).
Where is this apparent 16MiB initramfs size limit coming from?
There isn't one. A ramfs is only constrained by available RAM.
There is a definition for the "Default RAM disk size", but this would not affect the size of an initramfs.
Your method of booting with the U-Boot bootm command is suspect, i.e. you're passing the memory address of the initramfs as the second argument.
The U-Boot documentation clearly describes the second argument as "the address of an initrd image" (emphasis added).
There is no mention of initramfs as an argument.
Linux kernel documentation states that the initramfs archive can be "linked into the linux kernel image". There's a kernel menuconfig entry for specifying the path to the initramfs cpio file. The make script will append this cpio file to the kernel image so that there is a single image file for booting.
Or (like an initrd) "a separate file" can be passed to the kernel at boot to populate the initramfs.
U-Boot passes the location (and length) of this archive either as an ATAG entry or as a reserved memory region in the Device Tree blob.
The kernel expects either a cpio archive for the initramfs or a tar archive for an initrd.
You neglect to mention (besides its compression) what kind of archive this "initramfs" or "separate file" actually is .
So it's not clear if you're booting the kernel with an initrd (tar archive) or an initramfs (cpio archive).
Your repeated reference to the initramfs as an "image" file rather than a cpio archive suggests that you really are using an initrd.
An initrd would definitely have a size constraint.
Is there such a thing as an "kernel __init section" which is reclaimed by the kernel after it has uncompressed/loaded images?
Yes, there is an __init section of memory, which is released after kernel initialization is complete.
If so, how do I see/configure the location/size of it?
Usually routines and data that have no use after initialization can be declared with the __init macro. See this.
The location and size of this memory section would be under the control of the linker script, rather than explicit user control. The kernel System.map file should have the info for review.
I think you need to pass ramdisk_size in uboot bootargs command
ramdisk_size needs to be set if the ram-disk uncompress file
size is bigger than default setting. It should be more than
ramdisk uncompress file size.
While building the kernel I am giving LOADADDR as "0x80008000":
make uImage LOADADDR=0x80008000
Can you please help to understand what is the use of this? Can I change the LOADADDR, is there any restriction on the length of the LOADADDR?
(I'm assuming that you're using ARM based on the mention of U-Boot and the value of LOADADDR.)
Can you please help to understand what is the use of this?
LOADADDR specifies the address where the kernel image will be located by the linker. (This is true for a few architectures (e.g. Blackfin), but not for ARM.
LOADADDR specifies the address where the kernel image will be located by U-Boot and is stored in the U-Boot header by the mkimage utility. Typically the load address (for placement in memory) is also the start address (for execution). Note that the uImage file is typically just the (self-extracting, compressed) zImage file with the U-Boot wrapper.
Can I change the LOADADDR,
Yes, but according to (Vincent Sanders') Booting ARM Linux that would be contrary to ARM convention:
Despite the ability to place zImage anywhere within memory,
convention has it that it is loaded at the base of physical RAM plus
an offset of 0x8000 (32K). This leaves space for the parameter block
usually placed at offset 0x100, zero page exception vectors and page
tables. This convention is very common.
(The uImage mentioned in your question is probably just a zImage with the U-Boot wrapper, so the quotation does apply.)
is there any restriction on the length of the LOADADDR?
The "length"? If you're using a 32-bit processor, then the length of this address would be 32 bits.
ADDENDUM
arch/arm/boot/Makefile only uses LOADADDR for building the uImage from the zImage.
From (Russel King's) Booting ARM Linux the constraints on this LOADADDR are:
The
kernel should be placed in the first 128MiB of RAM. It is recommended
that it is loaded above 32MiB in order to avoid the need to relocate
prior to decompression, which will make the boot process slightly
faster.
When booting a raw (non-zImage) kernel the constraints are tighter.
In this case the kernel must be loaded at an offset into system equal
to TEXT_OFFSET - PAGE_OFFSET.
The expected locations for the Device Tree or ATAGs or an initramfs can add more constraints on this LOADADDR.
Putting this in other terms, kernel compilation can generate different images, compressed or not (i.e. needed for XIP).
LOADADDR is the entry_point of the kernel, must be correct (match with where u-boot loads it in DDR) for certain architectures that compiles with absolute long jumps, and not important at all for ARM that can execute relative jumps.
I am learning the linux kernel booting process and trying to install linux on my beagleboard xM. I came across two approach both using the SD card.
1. Have the MLO, initrd, uboot.bin and uImage in one partition.
2. Have the MLO, uboot.bin and uImage in one partition and a pre-built rootfs (Angstrom) in 2nd partition.
In the first approach how is the initrd transformed into a complete file system on the 2nd partition. And what happens internally when uboot extracts the kernel from uImage and paste it on the 2nd partition. which directories would get modified of the init rootfs. How the init process of the kernel is called.
In general, the booting sequence in the Beagleboard-XM is something like this :
Firmware -> MLO -> uboot.bin -> uImage -> initrd(optional) -> rootfs
The initrd itself is a simple RAM based file system. It is generally used to mount the actual file system. Some of the modules which are required to mount the rootfs are bundled up into the initrd. Once the rootfs is mounted, the initramfs is cleaned up. If the features that are required to mount the rootfs is built into the kernel (uImage) itself, then there is no need of initrd. As I said earlier, initrd is a simple RAM based file system. That means if you use this as your filesystem, anything you write onto it will not be preserved. A few embedded applications actually use this as their rootfs.
Now answering to your question,
How is the initrd transformed into a complete file system on the 2nd partition?
Initially u-boot loads the kernel(uImage), initrd onto RAM and pass the appropriate the command line arguments to the kernel. Once the u-boot transfers the control over to the kernel, it first initializes the drivers and other modules and finally looks for initrd. The initrd is generally put as a cpio image. The kernel then uncompresses it and looks for 'init'. The init processes few things and then mounts the rootfs. There is no such thing as 2nd partition when it comes to initrd. It resides on the RAM.
What happens internally when uboot extracts the kernel from uImage and paste it on the 2nd partition?
Again, there is no such things as 2nd partition when it comes to uImage. uImage is a format which is understood by u-boot. The u-boot first loads the uImage onto the RAM and before actually passing the control over to the kernel, it extracts the contents from uImage, moves the binary to a predefined location in the RAM and then passes the control over to the kernel. Similar to initrd, the kernel also resides on the RAM itself.
Which directories would get modified of the init rootfs?
No directories gets modifies. But yes, a few of them are mounted, like procfs, sysfs, etc.
How the init process of the kernel is called?
Please refer to the kernel_init() function under init/main.c to see how the kernel executes init (refer http://lxr.free-electrons.com/source/init/main.c#L871).
I am attaching a simple code snippet from it :
if (!try_to_run_init_process("/sbin/init") ||
!try_to_run_init_process("/etc/init") ||
!try_to_run_init_process("/bin/init") ||
!try_to_run_init_process("/bin/sh"))
return 0;
panic("No working init found. Try passing init= option to kernel. "
"See Linux Documentation/init.txt for guidance.");
Therefore, the kernel searches in these default paths for the init binary. If it is not found, the kernel panics!
Hope this helps.
I am working on Freescale board imx50evk. I have built the uboot.bin and uImage using LTIB (linux target image builder). At the U-Boot prompt I enter the bootm addr command, and then it hangs after showing the message "Loading Kernel..."
> MX50_RDP U-Boot > boot
MMC read: dev # 0, block # 2048, count 6290 partition # 0 ...
6290 blocks read: OK
## Booting kernel from Legacy Image at 70800000 ...
Image Name: Linux-2.6.35.8
Image Type: ARM Linux Kernel Image (uncompressed)
Data Size: 1323688 Bytes = 1.3 MB
Load Address: a0008000
Entry Point: a0008000
Verifying Checksum ... OK
Loading Kernel Image ...
You need to verify that your board really has RAM at 0xa0008000, which is the kernel "load address". U-Boot is probably trying to copy the image to that region of memory when it appears to hang.
[By your comment, I'll assume that you have verified that main memory does not exist at physical address 0xAXXXXXXX.]
The uImage file that you are using was made from the zImage file using the mkimage utility.
You probably have to manually edit the line that looks like
zreladdr-y := 0xa0008000
in arch/arm/mach-XXX/Makefile.boot for your board. The convention is that this address should be the base of physical RAM plus an offset of 0x8000 (32K). Then adjust the other values in the file. Delete the zImage file and perform another make for the kernel.
While building 3.20 development kernels for rockchip's rk3288 I found using LZO image compression made the device hang at 'Starting the kernel.' I assume it's because of a version miss-match between the build hosts LZO and the deployed decompression code, so it could probably happen with any of the compression algorithms. In my case switch to gzip fixed it.
This is only my assumption for why changing the compression algorithm gave a bootable kernel. Please correct me if I'm wrong.