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linux – 为什么我不应该在ARMv6的系统内存上使用ioremap?

发布时间:2020-12-13 23:53:19 所属栏目:Linux 来源:网络整理
导读:我需要从内核中保留一个物理连续RAM的大缓冲区,并且能够保证缓冲区将始终使用特定的硬编码物理地址.此缓冲区应保留为内核的整个生命周期.我编写了一个chardev驱动程序作为用户空间中访问此缓冲区的接口.我的平台是一个嵌入式系统,ARMv7架构运行2.6 Linux内核
我需要从内核中保留一个物理连续RAM的大缓冲区,并且能够保证缓冲区将始终使用特定的硬编码物理地址.此缓冲区应保留为内核的整个生命周期.我编写了一个chardev驱动程序作为用户空间中访问此缓冲区的接口.我的平台是一个嵌入式系统,ARMv7架构运行2.6 Linux内核.

有关该主题(第443页)的第15章(共Linux Device Drivers,Third Edition页):

Reserving the top of RAM is accomplished by passing a mem= argument to the kernel at boot time. For example,if you have 256 MB,the argument mem=255M keeps the kernel from using the top megabyte. Your module could later use the following code to gain access to such memory:
dmabuf = ioremap (0xFF00000 /* 255M */,0x100000 /* 1M */);

我已经完成了这个以及其他一些事情:

>除了mem之外,我还在使用memmap bootarg. kernel boot parameters documentation建议在使用mem时始终使用memmap以避免地址冲突.
>我在调用ioremap之前使用了request_mem_region,当然,在继续前进之前,我检查它是否成功.

这是我完成所有这些后系统的样子:

# cat /proc/cmdline 
root=/dev/mtdblock2 console=ttyS0,115200 init=/sbin/preinit earlyprintk debug mem=255M memmap=1M$255M
# cat /proc/iomem 

  

  

Reserving the top of RAM is accomplished by passing a mem= argument to the kernel at boot time. For example,the argument mem=255M keeps the kernel from using the top megabyte. Your module could later use the following code to gain access to such memory: dmabuf = ioremap (0xFF00000 /* 255M */,0x100000 /* 1M */);

000-0fffffff : PCIe Outbound Window,Port 0

Reserving the top of RAM is accomplished by passing a mem= argument to the kernel at boot time. For example,0x100000 /* 1M */);

000-082fffff : PCI Bus 0001:01

Reserving the top of RAM is accomplished by passing a mem= argument to the kernel at boot time. For example,0x100000 /* 1M */);

000-081fffff : 0001:01:00.0 08200000-08207fff : 0001:01:00.0 18000300-18000307 : serial 18000400-18000407 : serial 1800c000-1800cfff : dmu_regs 18012000-18012fff : pcie0 18013000-18013fff : pcie1 18014000-18014fff : pcie2 19000000-19000fff : cru_regs 1e000000-1fffffff : norflash 40000000-47ffffff : PCIe Outbound Window,Port 1 40000000-403fffff : PCI Bus 0002:01 40000000-403fffff : 0002:01:00.0 40400000-409fffff : PCI Bus 0002:01 40400000-407fffff : 0002:01:00.0 4

Reserving the top of RAM is accomplished by passing a mem= argument to the kernel at boot time. For example,0x100000 /* 1M */);

00-40807fff : 0002:01:00.0 80000000-8fefffff : System RAM 80052000-8045dfff : Kernel text 80478000-80500143 : Kernel data 8ff00000-8fffffff : foomem=mem=255Mdmabuf = ioremap (0xFF00000 /* 255M */,0x100000 /* 1M */);mem=mem=255Mdmabuf = ioremap (0xFF00000 /* 255M */,0x100000 /* 1M */);

到目前为止,一切看起来都很好,我的司机工作得很好.我能够直接读取和写入我选择的特定物理地址.

但是,在启动过程中,触发了一个很大的可怕警告(?):

BUG: Your driver calls ioremap() on system memory.  This leads
to architecturally unpredictable behaviour on ARMv6+,and ioremap()
will fail in the next kernel release.  Please fix your driver.
------------[ cut here ]------------
WARNING: at arch/arm/mm/ioremap.c:211 __arm_ioremap_pfn_caller+0x8c/0x144()
Modules linked in:
[] (unwind_backtrace+0x0/0xf8) from [] (warn_slowpath_common+0x4c/0x64)
[] (warn_slowpath_common+0x4c/0x64) from [] (warn_slowpath_null+0x1c/0x24)
[] (warn_slowpath_null+0x1c/0x24) from [] (__arm_ioremap_pfn_caller+0x8c/0x144)
[] (__arm_ioremap_pfn_caller+0x8c/0x144) from [] (__arm_ioremap_caller+0x50/0x58)
[] (__arm_ioremap_caller+0x50/0x58) from [] (foo_init+0x204/0x2b0)
[] (foo_init+0x204/0x2b0) from [] (do_one_initcall+0x30/0x19c)
[] (do_one_initcall+0x30/0x19c) from [] (kernel_init+0x154/0x218)
[] (kernel_init+0x154/0x218) from [] (kernel_thread_exit+0x0/0x8)
---[ end trace 1a4cab5dbc05c3e7 ]---

触发自:arc/arm/mm/ioremap.c

/*
 * Don't allow RAM to be mapped - this causes problems with ARMv6+
 */
if (pfn_valid(pfn)) {
    printk(KERN_WARNING "BUG: Your driver calls ioremap() on system memory.  This leadsn"
           KERN_WARNING "to architecturally unpredictable behaviour on ARMv6+,and ioremap()n"
           KERN_WARNING "will fail in the next kernel release.  Please fix your driver.n");
    WARN_ON(1);
}

确切地说,这会导致什么问题?他们能减轻吗?我有什么选择?

解决方法

So I’ve done exactly that,and it’s working.

提供内核命令行(例如/ proc / cmdline)和生成的内存映射(即/ proc / iomem)来验证这一点.

What problems,exactly,could this cause?

在系统内存上使用ioremap()的问题是,您最终会将冲突的属性分配给内存,从而导致“不可预测”的行为.
请参阅文章“ARM’s multiply-mapped memory mess”,该文章提供了您正在触发的警告的历史记录.

The ARM kernel maps RAM as normal memory with writeback caching; it’s also marked non-shared on uniprocessor systems. The ioremap() system call,used to map I/O memory for CPU use,is different: that memory is mapped as device memory,uncached,and,maybe,shared. These different mappings give the expected behavior for both types of memory. Where things get tricky is when somebody calls ioremap() to create a new mapping for system RAM.

The problem with these multiple mappings is that they will have differing attributes. As of version 6 of the ARM architecture,the specified behavior in that situation is “unpredictable.”

请注意,“系统内存”是由内核管理的RAM.
触发警告的事实表明您的代码正在为内存区域生成多个映射.

Can they be mitigated?

您必须确保您想要ioremap()的RAM不是“系统内存”,即由内核管理.
另见this answer.

附录

关注您的此警告是pfn_valid(pfn)返回TRUE而不是FALSE的结果.
基于您为2.6.37版提供的Linux交叉引用链接,
pfn_valid()只是返回结果

memblock_is_memory(pfn << PAGE_SHIFT);

这反过来只是返回结果

memblock_search(&memblock.memory,addr) != -1;

我建议破解内核代码,以便揭示冲突.
在调用ioremap()之前,将TRUE赋值给全局变量memblock_debug.
以下补丁应显示有关内存冲突的重要信息.
(memblock列表按基址排序,因此memblock_search()在此列表上执行二进制搜索,因此使用mid作为索引.)

static int __init_memblock memblock_search(struct memblock_type *type,phys_addr_t addr)
 {
         unsigned int left = 0,right = type->cnt;

         do {
                 unsigned int mid = (right + left) / 2;

                 if (addr < type->regions[mid].base)
                         right = mid;
                 else if (addr >= (type->regions[mid].base +
                                   type->regions[mid].size))
                         left = mid + 1;
-                else
+                else {
+                        if (memblock_debug)
+                                pr_info("MATCH for 0x%x: m=0x%x b=0x%x s=0x%xn",+                                                addr,mid,+                                                type->regions[mid].base,+                                                type->regions[mid].size);
                         return mid;
+                }
         } while (left < right);
         return -1;
 }

如果要查看所有内存块,请使用变量memblock_debug调用memblock_dump_all()为TRUE.

[有趣的是,这实际上是一个编程问题,但我们还没有看到你的任何代码.]

附录2

由于您可能正在使用ATAG(而不是设备树),并且您想要专用内存区域,请修复ATAG_MEM以反映这种较小的物理内存大小.
假设您对引导代码进行了零更改,ATAG_MEM仍然指定了完整的RAM,因此这可能是导致警告的系统内存冲突的根源.
见this answer about ATAGs和this related answer.

(编辑:李大同)

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