http://stackoverflow.com/questions/263215/do-any-languages-compilers-utilize-the-x86-enter-instruction-with-a-zero-ne#comment41311163_263215” Idonotexist 指出,“海合会>/em>利用上文所显示的精确辛迪-克斯一号支持职能。
但是,它没有使用<条码>ENTER指令。 相反,用于加固功能的变数在地方变数领域被组合在一起,而这一组的一个点被移至封顶功能。 有趣的是,这一“母体变量的点子”是通过非标准机制通过的: 关于x64,在<代码>r10上通过,在x86(cdecl)上通过,保留在C++的<代码>this点(在任何方面不支持封闭功能)。
#include <stdio.h>
void func_a(void)
{
int a1 = 0x1001;
int a2=2, a3=3, a4=4;
int a5 = 0x1005;
void func_b(int p1, int p2)
{
/* Use variables from func_a() */
printf("a1=%d a5=%d
", a1, a5);
}
func_b(1, 2);
}
int main(void)
{
func_a();
return 0;
}
• 编制以下(不适用)代码,用于64比值:
00000000004004dc <func_b.2172>:
4004dc: push rbp
4004dd: mov rbp,rsp
4004e0: sub rsp,0x10
4004e4: mov DWORD PTR [rbp-0x4],edi
4004e7: mov DWORD PTR [rbp-0x8],esi
4004ea: mov rax,r10 ; ptr to calling function "shared" vars
4004ed: mov ecx,DWORD PTR [rax+0x4]
4004f0: mov eax,DWORD PTR [rax]
4004f2: mov edx,eax
4004f4: mov esi,ecx
4004f6: mov edi,0x400610
4004fb: mov eax,0x0
400500: call 4003b0 <printf@plt>
400505: leave
400506: ret
0000000000400507 <func_a>:
400507: push rbp
400508: mov rbp,rsp
40050b: sub rsp,0x20
40050f: mov DWORD PTR [rbp-0x1c],0x1001
400516: mov DWORD PTR [rbp-0x4],0x2
40051d: mov DWORD PTR [rbp-0x8],0x3
400524: mov DWORD PTR [rbp-0xc],0x4
40052b: mov DWORD PTR [rbp-0x20],0x1005
400532: lea rax,[rbp-0x20] ; Pass a, b to the nested function
400536: mov r10,rax ; in r10 !
400539: mov esi,0x2
40053e: mov edi,0x1
400543: call 4004dc <func_b.2172>
400548: leave
400549: ret
objdump-no-show-raw-insn> - d-Mintel
这相当于像以下几句:
struct func_a_ctx
{
int a1, a5;
};
void func_b(struct func_a_ctx *ctx, int p1, int p2)
{
/* Use variables from func_a() */
printf("a1=%d a5=%d
", ctx->a1, ctx->a5);
}
void func_a(void)
{
int a2=2, a3=3, a4=4;
struct func_a_ctx ctx = {
.a1 = 0x1001,
.a5 = 0x1005,
};
func_b(&ctx, 1, 2);
}
