对模拟代码中字节顺序的重要性感到非常困惑

Question

老实说，我无法理解这一点。我正在为 aarch64 生成代码，看来我可以运行 完全相同的代码 与 qemu-aarch64 和 qemu-aarch64_be 仅通过更改字节顺序ELF headers。否则，可执行文件 byte-for-byte 相同。在这两种情况下，Objdump 也能正确地反汇编代码。怎么可能呢？生成的代码（我认为）在磁盘上 little-endian，事实上，如果它是相反的字节顺序，它似乎不起作用。

Big-endian（headers 和前四个指令）：

00000000  7f 45 4c 46 02 02 01 00  00 00 00 00 00 00 00 00  |.ELF............|
00000010  00 02 00 b7 00 00 00 01  00 00 00 00 00 00 10 80  |................|
00000020  00 00 00 00 00 00 00 40  00 00 00 00 00 00 00 00  |.......@........|
00000030  00 00 00 00 00 40 00 38  00 01 00 00 00 00 00 00  |.....@.8........|
00000040  00 00 00 01 00 00 00 05  00 00 00 00 00 00 00 80  |................|
00000050  00 00 00 00 00 00 10 80  00 00 00 00 00 00 00 00  |................|
00000060  00 00 00 00 00 00 03 0c  00 00 00 00 00 00 03 0c  |................|
00000070  00 00 00 00 00 00 00 20  00 00 00 00 00 00 00 00  |....... ........|
00000080  ff 4f 00 d1 e0 03 40 39  00 20 00 11 e0 03 00 39  |.O....@9. .....9|

  80:   d1004fff    sub sp, sp, #0x13
  84:   394003e0    ldrb    w0, [sp]
  88:   11002000    add w0, w0, #0x8
  8c:   390003e0    strb    w0, [sp]

小端：

00000000  7f 45 4c 46 02 01 01 00  00 00 00 00 00 00 00 00  |.ELF............|
00000010  02 00 b7 00 01 00 00 00  80 10 00 00 00 00 00 00  |................|
00000020  40 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  |@...............|
00000030  00 00 00 00 40 00 38 00  01 00 00 00 00 00 00 00  |....@.8.........|
00000040  01 00 00 00 05 00 00 00  80 00 00 00 00 00 00 00  |................|
00000050  80 10 00 00 00 00 00 00  00 00 00 00 00 00 00 00  |................|
00000060  0c 03 00 00 00 00 00 00  0c 03 00 00 00 00 00 00  |................|
00000070  20 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  | ...............|
00000080  ff 4f 00 d1 e0 03 40 39  00 20 00 11 e0 03 00 39  |.O....@9. .....9|

  80:   d1004fff    sub sp, sp, #0x13
  84:   394003e0    ldrb    w0, [sp]
  88:   11002000    add w0, w0, #0x8
  8c:   390003e0    strb    w0, [sp]

我觉得我错过了什么。这里发生了什么？我唯一能想到的是 qemu 在幕后做了一些魔术，因为数据的布局方式显然没有任何效果。但是，如果是这样的话，它怎么知道字节序是正确的还是不正确的？

Answer 1

来自ARMv8-A Reference Manual：

B2.6.2  Instruction endianness

            In Armv8-A, A64 instructions have a fixed length of 32 bits
            and are always little-endian.

所以只有内存加载和存储受字节顺序影响。我仍然希望您的二进制文件以某种方式破坏在编译时初始化的数据，但从技术上讲，没有任何此类数据是可能的。

对模拟代码中字节顺序的重要性感到非常困惑

Very confused about the significance of endianness in emulated code

qemu

endianness

arm64

B2.6.2 Instruction endianness