从.so 文件到运行时 memory 的指令编码是否不同？

Question

I have a Android.so file uses ARM-V7A instruction set.

I drag it into IDA, and there is one line shows:

0x1000: b #0x102c

And the hex window show that the binary code of b #0x102c is 14 e0 .

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14 e0 have binrary format of 0001 0100 1110 0000 , which is not how ARM-manual encoded this instruction.

UNLESS

1    4    e    0
0001 0100 1110 0000
8 ----- 1 16 ---- 9

8 -- 1 means 1 bit to 8 bit, 9 -- 16 means 9 bit to 16bit

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Why is instruction encoded this way in .so file?

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For example if I want to runtime change some instruction at some address. Do I change it to 0x14e0 ( which is how instruction encoded in .so file ), or change it to 0xe014 ( which is how instrcution encoded in ARM-Manual )

Answer 1

Looks like IDA is breaking it into bytes so you'd expect the low 8 bits to be first, as a byte-stream. 14 e0 is the same thing as little-endian e014 , a single 16-bit halfword.

(IDA was probably first developed for x86, where machine code is a variable-length byte-stream, not a sequence of 16 or 32-bit chunks, and ported to ARM. This is still a valid way to hexdump ARM machine code, though.)

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Re: the title question:

The.text section of an executable or library will get mapped into memory like mmap , without modifications. Except in rare cases of "text relocations" to fix up absolute addresses, eg in jump tables in .rodata or movw / movk in .text itself.

But position-independent code generally avoids that, because that's the whole point of being position independent. It's definitely not byte-swapped!