ARM汇编 .word 解析

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转自:http://guliqun1983.blog.163.com/blog/static/5011168520111257719482/

3. 令人头疼的ARM汇编伪指令 .word经常碰到那些以“.”打头的一些令人头疼的伪指令,至于.globl _start .balign .align .data .text等等就算了,最最bt的如下:_undefined_instruction: .word undefined_instruction这个.word令人费解。网上的技术人员都不屑回答,说请参考GNU ASM。我去看了,对于.word解释如下:http://tigcc.ticalc.org/doc/gnuasm.html#SEC49.wordSyntax:
.word expressions This directive expects zero or more expressions, of any section, separated by commas. For each expression, as emits a 16-bit number for this target.以及as.info文档:7.92 .word expressionsThis directive expects zero or more expressions, of any
section, separated by commas.The size of the number emitted, and its byte order, depend on what target computerthe assembly is for.Warning: Special Treatment to support CompilersMachines with a 32-bit address space, but that do less than 32-bit addressing,
requirethe following special treatment. If the machine of interest to you does 32-bit addressing(or doesn’t require it; see Chapter 8 [Machine Dependencies], page 61), you can ignore thisissue.In order to assemble compiler output into something that works,
as occasionally doesstrange things to ‘.word’ directives. Directives of the form ‘.word sym1-sym2’ are oftenemitted by compilers as part of jump tables. Therefore, when as assembles a directive ofthe form ‘.word sym1-sym2’, and the difference between sym1
and sym2 does not fit in 16bits, as creates a secondary jump table, immediately before the next label. This secondaryjump table is preceded by a short-jump to the first byte after the secondary table. Thisshort-jump prevents the flow of control from accidentally
falling into the new table. Insidethe table is a long-jump to sym2. The original ‘.word’ contains sym1 minus the address ofthe long-jump to sym2.If there were several occurrences of ‘.word sym1-sym2’ before the secondary jump table,all of them are adjusted.
If there was a ‘.word sym3-sym4’, that also did not fit in sixteenbits, a long-jump to sym4 is included in the secondary jump table, and the .word directivesare adjusted to contain sym3 minus the address of the long-jump to sym4; and so on, for asmany entries
in the original jump table as necessary. 看了以后仍然一头雾水。我把bin文件反汇编,想通过这种方法来找找这个.word究竟干什么。原汇编程序:(start.S).globl _start_start: b reset ldr pc, _undefined_instruction ldr pc, _software_interrupt ldr pc, _prefetch_abort ldr pc, _data_abort ldr pc, _not_used ldr pc,
_irq ldr pc, _fiq _undefined_instruction: .word undefined_instruction_software_interrupt: .word software_interrupt_prefetch_abort: .word prefetch_abort_data_abort: .word data_abort_not_used: .word not_used_irq: .word irq_fiq: .word fiq .balignl 16,0xdeadbeef
_TEXT_BASE: .word TEXT_BASE .globl _armboot_start_armboot_start: .word _start .globl _bss_start_bss_start: .word __bss_start .globl _bss_end_bss_end: .word _end reset: /* * set the cpu to SVC32 mode */ mrs r0,cpsr bic r0,r0,#0x1f orr r0,r0,#0xd3 msr cpsr,r0
对应的反汇编:00000000 [0xea000012] b 0x5000000004 [0xe59ff014] ldr pc,0x00000020 ; = #0x33f8014000000008 [0xe59ff014] ldr pc,0x00000024 ; = #0x33f801a00000000c [0xe59ff014] ldr pc,0x00000028 ; = #0x33f8020000000010 [0xe59ff014] ldr pc,0x0000002c ; = #0x33f8026000000014
[0xe59ff014] ldr pc,0x00000030 ; = #0x33f802c000000018 [0xe59ff014] ldr pc,0x00000034 ; = #0x33f803200000001c [0xe59ff014] ldr pc,0x00000038 ; = #0x33f8038000000020 [0x33f80140] mvnccs r0,#0x10 ; ? rn = 0x800000024 [0x33f801a0] mvnccs r0,#0x28 ; ? rn = 0x800000028
[0x33f80200] mvnccs r0,#0, 4 ; ? rn = 0x80000002c [0x33f80260] mvnccs r0,#6 ; ? rn = 0x800000030 [0x33f802c0] mvnccs r0,#0xc ; ? rn = 0x800000034 [0x33f80320] mvnccs r0,#0x80000000 ; ? rn = 0x800000038 [0x33f80380] mvnccs r0,#2 ; ? rn = 0x80000003c [0xdeadbeef]
cdple p14,0xa,c11,c13,c15,700000040 [0x33f80000] mvnccs r0,#0 ; ? rn = 0x800000044 [0x33f80000] mvnccs r0,#0 ; ? rn = 0x800000048 [0x33f96650] mvnccs r6,#0x5000000 ; ? rn = 0x90000004c [0x33f9ab80] mvnccs r10,#0x20000 ; ? rn = 0x900000050 [0xe10f0000] mrs
r0,cpsr00000054 [0xe3c0001f] bic r0,r0,#0x1f00000058 [0xe38000d3] orr r0,r0,#0xd30000005c [0xe129f000] msr cpsr_cf,r0 这么看来,_undefined_instruction: .word undefined_instruction这句对应的反汇编是: mvnccs r0,#0x10 ; 这么一来我又更糊涂了。 到ChinaUnix求助。幸好碰到一位热心的网友wheelz,详细地给我解答了。帖子链接如下:http://www.linuxforum.net/forum/showflat.php?Cat=&Board=linuxK&Number=563178
现在总结wheelz的回答,说说这个.word的作用。 word expression就是在当前位置放一个word型的值,这个值就是expression 举例来说, _rWTCON: .word 0x15300000 就是在当前地址,即_rWTCON处放一个值0x15300000 翻译成intel的汇编语句就是: _rWTCON dw 0x15300000 就是在当前位置放个expression的值。 原来如此啊。 PS:贴一个##的作用。#define _syscall0(type,name) type
name(void) { long __res; __asm__ volatile ("int $0x80" : "=a" (__res) : "0" (__NR_##name)); if (__res >= 0) return (type) __res; errno = -__res; return -1; }__NR_##name是系统调用号,##指的是两次宏展开.即用实际的系统调用名字代替"name",然后再把__NR_...展开.如name == ioctl,则为__NR_ioctl。