问题描述
我目前正在学习 ARM 汇编语言;
I am currently learning ARM assembly language;
要做到这一点，我想将一些 86 code（AT＆amp; T公司语法）来ARM汇编（Intel语法） code。
To do so, I am trying to convert some x86 code (AT&T Syntax) to ARM assembly (Intel Syntax) code.
__asm__("movl $0x0804c000, %eax;");
__asm__("mov R0,#0x0804c000");
从这个，我知道在86的块1堆结构从0x0804c000开始。不过，我当我尝试做相同的手臂，
我收到以下错误：
From this document, I learn that in x86 the Chunk 1 of the heap structure starts from 0x0804c000. But I when I try do the same in arm, I get the following error:
/tmp/ccfNZp9F.s:174: Error: invalid constant (804c000) after fixup
我假设的问题是，ARM只能加载,我知道如何内存间接寻址的作品。
From my previous question, I know how memory indirect addressing works.
写下面做同样的工作的片段？
Are the snippets written below doing the same job?
movl (%eax), %ebx
LDR R0,[R1]
我使用的ARMv7处理器第4版（v7l）
推荐答案
问题1的答案
ARM的MOV指令仅具有可用于立即值12位，这些位被以这种方式使用：对值8位，和4位来指定转数到右边（转数乘以2，增加的范围内）。
The MOV instruction on ARM only has 12 bits available for an immediate value, and those bits are used this way: 8 bits for value, and 4 bits to specify the number of rotations to the right (the number of rotations is multiplied by 2, to increase the range).
这意味着，仅仅值的有限数量的可以与该指令一起使用。它们是：
This means that only a limited number of values can be used with that instruction. They are:
- 0-255
- 256，260，264，...，1020
- 1024，1040，1056，...，4080
- 等
等。你得到的错误，因为你常不能用8位+旋转来创建。您可以在值加载到寄存器下面的指令：
And so on. You are getting that error because your constant can't be created using the 8 bits + rotations. You can load that value onto the register following instruction:
LDR r0, =0x0804c000
请注意，这是一个伪指令，但。汇编程序将基本把那个地方不断在code和加载它与一些偏移量与PC（程序计数器）的内存位置。
Notice that this is a pseudo-instruction though. The assembler will basically put that constant somewhere in your code and load it as a memory location with some offset to the PC (program counter).
回答问题2
是那些指令是等价的。
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