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如何在 ARM64 Linux 中创建 1:1 MMU 映射？

[英]How to create 1:1 MMU mapping in ARM64 Linux?

原文 2021-03-14 11:18:14 4 1 linux/ virtual-memory/ arm64/ mmu

I'm trying to switch off MMU to run some test code.我正在尝试关闭 MMU 以运行一些测试代码。 I need to create exactly 1:1 virtual to physical memory mapping for code to run after MMU turned off.我需要创建精确的 1:1 虚拟到物理 memory 映射，以便在 MMU 关闭后运行代码。

In this question a guy described a method to create such a mapping with assembly code, but it seems like by the time he creates mapping, system is unfunctional. 在这个问题中，一个人描述了一种使用汇编代码创建这种映射的方法，但似乎在他创建映射时，系统已经无法正常工作。

My question is:我的问题是：

Is it possible to create a 1:1 MMU mapping, leaving linux functional (to test mapping and to load test code with devmem file)?是否可以创建 1:1 MMU 映射，让 linux 正常工作（测试映射并使用 devmem 文件加载测试代码）？
If possible, what is the best option to create it?如果可能，创建它的最佳选择是什么？

1 个解决方案

I need to create exactly 1:1 virtual to physical memory mapping for code to run after MMU turned off我需要创建精确的 1:1 虚拟到物理 memory 映射，以便在 MMU 关闭后运行代码

That simply does not make sense.这根本没有意义。 Memory mapping is used only when MMU is ON . Memory 映射仅在 MMU 为ON时使用。

Once MMU is off there is no virtual-to-physical translations and memory is access by physical addresses.一旦 MMU 关闭，就没有虚拟到物理的转换，并且 memory 可以通过物理地址访问。 Well in such way it could be described as '1:1 virtual to physical'.嗯，以这种方式，它可以被描述为“1:1 虚拟到物理”。

Problem with switching off MMU is that code which does that should be specially build to run with that 1:1 mapping.关闭 MMU 的问题是应该专门构建以使用该 1:1 映射运行的代码。

For example code is build to call function with virtual address 0x8000 .例如，构建代码以使用虚拟地址0x8000调用 function 。 OS is placing you program at physical address 0x00C0_0000 .操作系统将您的程序放置在物理地址0x00C0_0000 。 So basically jump into 0x8000 is jumping to physical 0x00C0_8000 .所以基本上跳转到0x8000就是跳转到物理0x00C0_8000 。
Once MMU is off, jump into 0x8000 would be into physical 0x8000 .一旦 MMU 关闭，跳转到0x8000将进入物理0x8000 。