我一直在研究如何通過拆解C代碼在x86架構中處理浮點操作。使用的操作系統是一個64位的Linux,而代碼是爲32位機器編譯的。32位架構的DS段寄存器
這裏是C源代碼:
#include <stdio.h>
#include <float.h>
int main(int argc, char *argv[])
{
float a, b;
float c, d;
printf("%u\n",sizeof(float));
a = FLT_MAX;
b = 5;
c = a/b;
d = (float) a/(float) b;
printf("%f %f \n",c,d);
return 0;
}
這裏是32位的EXE的主要功能的反彙編版本:
804841c: 55 push ebp
804841d: 89 e5 mov ebp,esp
804841f: 83 e4 f0 and esp,0xfffffff0
8048422: 83 ec 30 sub esp,0x30
8048425: c7 44 24 04 04 00 00 mov DWORD PTR [esp+0x4],0x4
804842c: 00
804842d: c7 04 24 20 85 04 08 mov DWORD PTR [esp],0x8048520
8048434: e8 b7 fe ff ff call 80482f0 <[email protected]>
8048439: a1 2c 85 04 08 mov eax,ds:0x804852c
804843e: 89 44 24 2c mov DWORD PTR [esp+0x2c],eax
8048442: a1 30 85 04 08 mov eax,ds:0x8048530
8048447: 89 44 24 28 mov DWORD PTR [esp+0x28],eax
804844b: d9 44 24 2c fld DWORD PTR [esp+0x2c]
804844f: d8 74 24 28 fdiv DWORD PTR [esp+0x28]
8048453: d9 5c 24 24 fstp DWORD PTR [esp+0x24]
8048457: d9 44 24 2c fld DWORD PTR [esp+0x2c]
804845b: d8 74 24 28 fdiv DWORD PTR [esp+0x28]
804845f: d9 5c 24 20 fstp DWORD PTR [esp+0x20]
8048463: d9 44 24 20 fld DWORD PTR [esp+0x20]
8048467: d9 44 24 24 fld DWORD PTR [esp+0x24]
804846b: d9 c9 fxch st(1)
804846d: dd 5c 24 0c fstp QWORD PTR [esp+0xc]
8048471: dd 5c 24 04 fstp QWORD PTR [esp+0x4]
8048475: c7 04 24 24 85 04 08 mov DWORD PTR [esp],0x8048524
804847c: e8 6f fe ff ff call 80482f0 <[email protected]>
8048481: b8 00 00 00 00 mov eax,0x0
8048486: c9 leave
8048487: c3 ret
8048488: 66 90 xchg ax,ax
804848a: 66 90 xchg ax,ax
804848c: 66 90 xchg ax,ax
804848e: 66 90 xchg ax,ax
什麼我有麻煩理解是行哪裏浮點值被傳送到寄存器。具體做法是:
mov eax,ds:0x804852c
mov eax,ds:0x8048530
在我的理解,所述指令應等於MOV EAX,[0x804852c]和MOV EAX,[0x8048530]分別因爲在32位模式ds寄存器通常指向整個32位的空間和通常是0。但是當我檢查寄存器值的ds是不爲0,具有
ds 0x2b
鑑於值,不應計算是
0x2b *0x10 + 0x8048520
然而FLOA值ts存儲在0x8048520和0x8048530中,就像DS中的值爲0一樣。任何人都可以向我解釋這是爲什麼?
http://stackoverflow.com/questions/3819699/what-does-ds40207a-mean-in-assembly –