Intel x86到ARM汇编转换(Intel x86 to ARM assembly conversion)

 我目前正在学习ARM汇编语言; 
 
 为此，我试图将一些x86 code (AT&T Syntax)为ARM汇编(Intel Syntax)代码。 
 
__asm__("movl $0x0804c000, %eax;");

__asm__("mov R0,#0x0804c000");
 
 从这个文档中 ，我了解到在x86中，堆结构的Chunk 1从0x0804c000开始。 但是，当我尝试做同样的操作时，我收到以下错误： 
 
/tmp/ccfNZp9F.s:174: Error: invalid constant (804c000) after fixup
 
 我假设问题是ARM只能加载32位指令。 
 
Question 1: Any idea what would be the first chunk in case of ARM processors?


Question 2:
 
 从我之前的问题 ，我知道内存间接寻址是如何工作的。 
 
 下面写的片段是否做同样的工作？ 
 
movl (%eax), %ebx

LDR R0,[R1]
 
 我正在使用ARMv7 Processor rev 4 (v7l) 

I am currently learning ARM assembly language; 
 
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");
 
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
 
I am assuming the problem is that ARM can only load 32bit instructions. 
 
Question 1: Any idea what would be the first chunk in case of ARM processors?


Question 2:
 
From my previous question, I know how memory indirect addressing works. 
 
Are the snippets written below doing the same job?
 
movl (%eax), %ebx

LDR R0,[R1]
 
I am using ARMv7 Processor rev 4 (v7l)

原文：https://stackoverflow.com/questions/17297695