为什么 AT&T jmp *bar 组装成带有 SIB 字节的机器代码?指令中没有使用寄存器 [英] Why does AT&T jmp *bar assemble to machine code with a SIB byte? There's no register used in the instruction

问题描述

我在 x86 机器上有这个绝对间接跳转指令:

I have this absolute indirect jump instruction on x86 machine:

ff 24 25 30 10 60 00

产生于:

jmp *bar

但是我在解码它的第二个和第三个字节时遇到了麻烦.

but I've got trouble with decoding second and third byte of it.

第二个应该是 Mod R/M 字段.所以它转化为:

Second should be the Mod R/M field. So it translates to:

00 100 100

含义:

00 - 没有位移的内存(但它有常量地址，这不是位移"吗?)

00 - memory with no displacement (but it has the constant address, isn't this "displacement"?)

100 (dec. 4) - 扩展操作.代码 (FF/4 => JMP r/m32)

100 (dec. 4) - extended op. code (FF/4 => JMP r/m32)

100 - ??SIB?但是这个指令没有用到寄存器

100 - ?? SIB? but there is no register used in this instruction

附言一些背景:

Breakpoint 4, test () at test.s:13
13              jmp     *bar
(gdb) disassemble /r
Dump of assembler code for function test:
   0x000000000040051b <+0>:     c7 04 25 30 10 60 00 2f 05 40 00        movl   $0x40052f,0x601030
=> 0x0000000000400526 <+11>:    ff 24 25 30 10 60 00    jmpq   *0x601030
   0x000000000040052d <+18>:    87 c0   xchg   %eax,%eax
   0x000000000040052f <+20>:    c3      retq   
End of assembler dump.
(gdb) list
8               bar: .word 0x0
9       .text
10      test:
11      .LFB0:
12              movl    $label1, bar
13              jmp     *bar
14              xchg    %eax, %eax
15              label1:
16              ret
17      .LFE0:
(gdb) 

推荐答案

操作数字节 24 25 表示具有 32 位位移的绝对寻址模式，即没有基址寄存器和索引寄存器.在 64 位模式下，这与 modr/m 字节 25 表示的 32 位位移的相对寻址模式不同(但在 32 位模式下不是).如果你想要后一种寻址方式，用AT&T语法写jmp *bar(%rip).

The operand bytes 24 25 indicate an absolute addressing mode with a 32 bit displacement, i.e. no base register and no index register. In 64 bit mode, this is distinct from the relative addressing mode with 32 bit displacement indicated by modr/m byte 25 (but in 32 bit mode it is not). If you want the latter addressing mode, write jmp *bar(%rip) in AT&T syntax.

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