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大家好,在学习位段的时候看到这段代码objdump的结果,其中很是不解。
代码如下:
struct {
short a:4;
short b:5;
short c:7;
} t;
int main()
{
t.a = 1;
t.b = 2;
t.c = 3;
printf(“size: %d\n”, sizeof(t));
printf(%d\n”, t.a);
printf(%d\n”, t.b);
printf(%d\n”, t.c);
}
下面是别人objdump的结果和一些注释:
t.a = 1; 是这样的 <<<<<<<<<<如何知道下面这段汇编是t.a=1产生
80483a0: 0f b7 05 d4 96 04 08 movzwl 0x80496d4,%eax
80483a7: 66 89 45 e8 mov %ax,0xffffffe8(%ebp)
80483ab: 8b 45 e8 mov 0xffffffe8(%ebp),%eax
80483ae: 83 e0 f0 and $0xfffffff0,%eax
80483b1: 83 c8 01 or $0x1,%eax
80483b4: 66 89 45 e8 mov %ax,0xffffffe8(%ebp)
80483b8: 8b 45 e8 mov 0xffffffe8(%ebp),%eax
80483bb: 66 a3 d4 96 04 08 mov %ax,0x80496d4
1、 movzwl 0x80496d4, %eax
取得 t 值放在 eax 中,在这里可以看出,sizeof(t)为2, 也就是 word size
2、 and $0xfffffff0, %eax
or $0x1, %eax
保留低4位, 然后置为1
3、 mov %ax, 0x80496d4
回写 t 值,t.a 定义为4个位,经过这3步,从而使 t 的低4位置为1。
下面是我自己尝试objdump的结果:
[root@localhost c_work]# objdump -d bit.o
bit.o: file format elf32-i386
Disassembly of section .text:
00000000 <main>:
0: 8d 4c 24 04 lea 0x4(%esp),%ecx
4: 83 e4 f0 and $0xfffffff0,%esp
7: ff 71 fc pushl 0xfffffffc(%ecx)
a: 55 push %ebp
b: 89 e5 mov %esp,%ebp
d: 51 push %ecx
e: 83 ec 14 sub $0x14,%esp
11: 0f b6 05 00 00 00 00 movzbl 0x0,%eax
18: 83 e0 f0 and $0xfffffff0,%eax
1b: 83 c8 01 or $0x1,%eax
1e: a2 00 00 00 00 mov %al,0x0
23: 0f b7 05 00 00 00 00 movzwl 0x0,%eax
2a: 66 25 0f fe and $0xfe0f,%ax
2e: 83 c8 20 or $0x20,%eax
31: 66 a3 00 00 00 00 mov %ax,0x0
37: 0f b6 05 01 00 00 00 movzbl 0x1,%eax
3e: 83 e0 01 and $0x1,%eax
41: 83 c8 06 or $0x6,%eax
44: a2 01 00 00 00 mov %al,0x1
49: c7 44 24 04 04 00 00 movl $0x4,0x4(%esp)
50: 00
51: c7 04 24 00 00 00 00 movl $0x0,(%esp)
58: e8 fc ff ff ff call 59 <main+0x59>
5d: 0f b6 05 00 00 00 00 movzbl 0x0,%eax
64: c1 e0 04 shl $0x4,%eax
67: c0 f8 04 sar $0x4,%al
6a: 0f be c0 movsbl %al,%eax
6d: 89 44 24 04 mov %eax,0x4(%esp)
71: c7 04 24 0e 00 00 00 movl $0xe,(%esp)
78: e8 fc ff ff ff call 79 <main+0x79>
7d: 0f b7 05 00 00 00 00 movzwl 0x0,%eax
84: c1 e0 07 shl $0x7,%eax
87: 66 c1 f8 0b sar $0xb,%ax
8b: 0f be c0 movsbl %al,%eax
8e: 89 44 24 04 mov %eax,0x4(%esp)
92: c7 04 24 16 00 00 00 movl $0x16,(%esp)
99: e8 fc ff ff ff call 9a <main+0x9a>
9e: 0f b6 05 01 00 00 00 movzbl 0x1,%eax
a5: d0 f8 sar %al
a7: 0f be c0 movsbl %al,%eax
aa: 89 44 24 04 mov %eax,0x4(%esp)
ae: c7 04 24 1e 00 00 00 movl $0x1e,(%esp)
b5: e8 fc ff ff ff call b6 <main+0xb6>
ba: 83 c4 14 add $0x14,%esp
bd: 59 pop %ecx
be: 5d pop %ebp
bf: 8d 61 fc lea 0xfffffffc(%ecx),%esp
c2: c3 ret
小弟敢问一下:
1 学好C,需要看懂objdump产生的代码吗?
2 如何看懂这些汇编代码?
谢谢 |
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