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cc編譯c語(yǔ)言中內(nèi)嵌匯編
--AT&T and Intel 匯編語(yǔ)法對(duì)照
寄存器命名：
AT&T: %eax
Intel: eax
AT&T 語(yǔ)法源地址在左側(cè)，目的地址在右側(cè)與Intel 方式語(yǔ)法相反
將eax值傳入ebx
AT&T: movl %eax, %ebx
Intel: mov ebx, eax
AT&T 語(yǔ)法在立即數(shù)前有前綴$.
AT&T: movl $0x0h, %eax
Intel: mov eax,0x0h
AT&T 語(yǔ)法在操作符后跟表示操作數(shù)類型的后綴b,w,l分別表示字節(jié)，字，雙字，相當(dāng)于偽操作符ptr，如果不加的話GAS會(huì)guess
AT&T: movw %ax, %bx
Intel: mov bx, ax
內(nèi)存尋址方式
AT&T: immed32(basepointer,indexpointer,indexscale)
Intel: [basepointer + indexpointer*indexscale + immed32]
地址計(jì)算公式為：
immed32 + basepointer + indexpointer * indexscale
直接尋址
AT&T: _a
Intel: [_a]
間接尋址
AT&T: (%eax)
Intel: [eax]
相對(duì)尋址
AT&T: _variable(%eax)
Intel: [eax + _variable]
AT&T: _array(,%eax,4)
Intel: [eax*4 + array]
C 代碼: *(p+1) p定義為char *
AT&T: 1(%eax) where eax has the value of p
Intel: [eax + 1]
結(jié)構(gòu)體數(shù)組尋址，結(jié)構(gòu)體長(zhǎng)度為8，下標(biāo)存于eax，結(jié)構(gòu)體內(nèi)偏移地址存于ebx，_array為結(jié)構(gòu)體數(shù)組首地址
AT&T: _array(%ebx,%eax,8)
Intel: [ebx + eax*8 + _array]
函數(shù)內(nèi)部實(shí)現(xiàn)交換
1、輸入與輸出變量相同
匯編代碼部分標(biāo)準(zhǔn)的交換實(shí)現(xiàn)，輸入部分用0寄存器表示"=r"(a)中所指定的寄存器即輸入與輸出變量相同
int main()
{ 804842c: mov 0xfffffff4(%ebp),%ecx
int a = 10, b = 0; 804842f: mov 0xfffffff0(%ebp),%edx
printf("before swap: a = %2d, b = %2d\n", a , b); 8048432: mov %ecx,%ebx
__asm__("nop; 8048434: mov %edx,%esi
movl %0, %%eax; 8048436: nop
movl %1, %0; 8048437: mov %ebx,%eax
movl %%eax, %1; 8048439: mov %esi,%ebx
nop;" 804843b: mov %eax,%esi
: 804843d: nop
"=r"(a), "=r"(b) 804843e: mov %ebx,%edx
: 8048440: mov %esi,%ecx
"0"(a), "1"(b) 8048442: mov %edx,%eax
: 8048444: mov %eax,0xfffffff4(%ebp)
"%eax" 8048447: mov %ecx,%eax
); 8048449: mov %eax,0xfffffff0(%ebp)
printf("after swap: a = %2d, b = %2d\n", a, b);
return 0;
}
2、輸入與輸出用不同的寄存器，&表示輸入輸出需要分配不同的寄存器
int main()
{
int a = 10, b = 0;
printf("before swap: a = %2d, b = %2d\n", a, b); 804842b: mov 0xfffffff8(%ebp),%edx
__asm__("nop; 804842e:
mov 0xfffffff4(%ebp),%eax
movl %2,
%1; 8048431:
nop
movl %3,
%0; 8048432:
mov %edx,%ebx
nop;"
8048434: mov %eax,%ecx
:
8048436: nop
"=&r"(a), "=&r"(b)
8048437: mov %ecx,%eax
:
8048439: mov %ebx,%edx
"r"(a), "r"(b)
804843b: mov %eax,%eax
);
804843d: mov %eax,0xfffffff8(%ebp)
printf("after swap: a = %2d, b = %2d\n", a , b); 8048440: mov %edx,%eax
return
0;
8048442: mov %eax,0xfffffff4(%ebp)
}
3、交換函數(shù)，需要間接尋址
#include <stdio.h>
void swap(int* x, int* y) 08048400 <swap>:
{ 8048400: push %ebp
__asm__("nop; 8048401: mov %esp,%ebp
movl (%0), %%eax; 8048403: push %ebx
movl (%1), %%ebx; 8048404: mov 0x8(%ebp),%ecx
movl %%ebx, (%0); 8048407: mov 0xc(%ebp),%edx
movl %%eax, (%1); 804840a: nop
nop;" 804840b: mov (%ecx),%eax
: 804840d: mov (%edx),%ebx
: 804840f: mov %ebx,(%ecx)
"r"(x),"r"(y) 8048411: mov %eax,(%edx)
: 8048413: nop
"eax", "ebx", "memory" 8048414: mov (%esp,1),%ebx ;ebx還原
); 8048417:
leave ;movl %ebp, %esp; pop ebp
} 8048418: ret
8048419: lea 0x0(%esi),%esi
int main()
{
int a = 10, b = 0;
printf("before swap: a = %2d, b = %2d\n", a, b);
swap(&a, &b);
printf("after swap: a = %2d, b = %2d\n", a, b);
return 0;
}
4、從匯編代碼中分離函數(shù)
1> 獲得匯編代碼
這里用加法函數(shù)，源代碼為：
int sum(int a, int b)
{
int c = a + b;
return c;
}
對(duì)應(yīng)的匯編代碼為
Disassembly of section .text:
00000000 <sum>:
0: 55 push %ebp
1: 89 e5 mov %esp,%ebp
3: 83 ec 04 sub $0x4,%esp
6: 8b 45 0c mov 0xc(%ebp),%eax
9: 03 45 08 add 0x8(%ebp),%eax
c: 89 45 fc mov %eax,0xfffffffc(%ebp)
f: 8b 45 fc mov 0xfffffffc(%ebp),%eax
12: 89 c0 mov %eax,%eax
14: c9 leave
15: c3 ret
16: 89 f6 mov %esi,%esi
2> 編寫內(nèi)嵌匯編語(yǔ)言函數(shù)
分析：為函數(shù)構(gòu)建運(yùn)行時(shí)堆棧情況即可使其順利運(yùn)行，由于編譯器在函數(shù)執(zhí)行開(kāi)始和結(jié)束時(shí)會(huì)增加
routine begin：
push %ebp; mov %esp, %ebp
routine end:
leave; ret
將上面的0, 1, 14, 15去掉，返回參數(shù)放在eax中將輸出部分設(shè)置為"=a"(r) 用eax寄存器 r 為需要的return type
步驟：
i 定義return_type r 變量
ii 去掉push %ebp; mov %esp, %ebp; leave; ret
iii 輸出部分為:"=a"(r):
$ vi sumassemble.c
int sum(int a, int b)
{
int r;
__asm__("sub $0x4, %%esp;
movl 0xc(%%ebp), %%eax;
addl 0x8(%%ebp), %%eax;
movl %%eax, 0xfffffffc(%%ebp);
movl 0xfffffffc(%%ebp), %%eax;
movl %%eax, %%eax;"
:
"=a"(r)
);
return r;
}
Disassembly of section .text:
00000000 <sum>:
0: 55 push %ebp
1: 89 e5 mov %esp,%ebp
3: 83 ec 04 sub $0x4,%esp
6: 83 ec 04 sub $0x4,%esp
9: 8b 45 0c mov 0xc(%ebp),%eax
c: 03 45 08 add 0x8(%ebp),%eax
f: 89 45 fc mov %eax,0xfffffffc(%ebp)
12: 8b 45 fc mov 0xfffffffc(%ebp),%eax
15: 89 c0 mov %eax,%eax
17: 89 c0 mov %eax,%eax
19: 89 45 fc mov %eax,0xfffffffc(%ebp)
1c: 8b 45 fc mov 0xfffffffc(%ebp),%eax
1f: 89 c0 mov %eax,%eax
21: c9 leave
22: c3 ret
23: 90 nop
3> 編譯可執(zhí)行程序
$ vi summain.c
extern int sum(int ,int);
int main()
{
int x = sum(1,2);
printf("x = %d\n", x);
return 0;
}
$ cc -o sum_main sum_main.c sum_assemble.c
$ ./sum_main
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