a[i]==i[a]==*(i+a)==*(a+i)

在C语言中，如果我们要访问一个数组的某个下标对应的元素,通常的写法是a[i]。但从汇编的角度看，写成i[a]一点问题都没有。

下面通过代码给出证明。

o foo1.c

 int main(int argc, char *argv[])
 {
     unsigned int a[] = {, , };
     unsigned int n = sizeof (a) / sizeof (int);

     unsigned int sum = ;
     for (unsigned int i = ; i < n; i++)
         sum += a[i];

     return sum;
 }

o foo2.c

 int main(int argc, char *argv[])
 {
     unsigned int a[] = {, , };
     unsigned int n = sizeof (a) / sizeof (int);

     unsigned int sum = ;
     for (unsigned int i = ; i < n; i++)
         sum += i[a];

     return sum;
 }

o foo3.c

 int main(int argc, char *argv[])
 {
     unsigned int a[] = {, , };
     unsigned int n = sizeof (a) / sizeof (int);

     unsigned int sum = ;
     for (unsigned int i = ; i < n; i++)
         sum += *(i+a);

     return sum;
 }

o 编译和运行

 $ gcc -g -Wall -std=gnu99 -m32 -o foo1 foo1.c
 $ gcc -g -Wall -std=gnu99 -m32 -o foo2 foo2.c
 $ gcc -g -Wall -std=gnu99 -m32 -o foo3 foo3.c
 $ ./foo1; echo $?

 $ ./foo2; echo $?

 $ ./foo3; echo $?
 

o 反汇编后diff

1) foo1.gdb.out

 (gdb) disas /m main
 Dump of assembler code for function main:
        {
    0x080483ed <+>:     push   ebp
    0x080483ee <+>:     mov    ebp,esp
    0x080483f0 <+>:     sub    esp,0x20

                unsigned int a[] = {, , };
    0x080483f3 <+>:     mov    DWORD PTR [ebp-0xc],0x1
    0x080483fa <+>:    mov    DWORD PTR [ebp-0x8],0x2
    0x08048401 <+>:    mov    DWORD PTR [ebp-0x4],0x3

                unsigned int n = sizeof (a) / sizeof (int);
    0x08048408 <+>:    mov    DWORD PTR [ebp-0x10],0x3

                unsigned int sum = ;
    0x0804840f <+>:    mov    DWORD PTR [ebp-0x18],0x0

                for (unsigned int i = ; i < n; i++)
    0x08048416 <+>:    mov    DWORD PTR [ebp-0x14],0x0
    0x0804841d <+>:    jmp    0x804842d <main+>
    0x08048429 <+>:    add    DWORD PTR [ebp-0x14],0x1
    0x0804842d <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x08048430 <+>:    cmp    eax,DWORD PTR [ebp-0x10]
    0x08048433 <+>:    jb     0x804841f <main+>

                        sum += a[i];
    0x0804841f <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x08048422 <+>:    mov    eax,DWORD PTR [ebp+eax*-0xc]
    0x08048426 <+>:    add    DWORD PTR [ebp-0x18],eax

               return sum;
    0x08048435 <+>:    mov    eax,DWORD PTR [ebp-0x18]

       }
    0x08048438 <+>:    leave
    0x08048439 <+>:    ret

 End of assembler dump.

2) foo2.gdb.out

 (gdb) disas /m main
 Dump of assembler code for function main:
        {
    0x080483ed <+>:     push   ebp
    0x080483ee <+>:     mov    ebp,esp
    0x080483f0 <+>:     sub    esp,0x20

                unsigned int a[] = {, , };
    0x080483f3 <+>:     mov    DWORD PTR [ebp-0xc],0x1
    0x080483fa <+>:    mov    DWORD PTR [ebp-0x8],0x2
    0x08048401 <+>:    mov    DWORD PTR [ebp-0x4],0x3

                unsigned int n = sizeof (a) / sizeof (int);
    0x08048408 <+>:    mov    DWORD PTR [ebp-0x10],0x3

                unsigned int sum = ;
    0x0804840f <+>:    mov    DWORD PTR [ebp-0x18],0x0

                for (unsigned int i = ; i < n; i++)
    0x08048416 <+>:    mov    DWORD PTR [ebp-0x14],0x0
    0x0804841d <+>:    jmp    0x804842d <main+>
    0x08048429 <+>:    add    DWORD PTR [ebp-0x14],0x1
    0x0804842d <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x08048430 <+>:    cmp    eax,DWORD PTR [ebp-0x10]
    0x08048433 <+>:    jb     0x804841f <main+>

                        sum += i[a];
    0x0804841f <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x08048422 <+>:    mov    eax,DWORD PTR [ebp+eax*-0xc]
    0x08048426 <+>:    add    DWORD PTR [ebp-0x18],eax

               return sum;
    0x08048435 <+>:    mov    eax,DWORD PTR [ebp-0x18]

       }
    0x08048438 <+>:    leave
    0x08048439 <+>:    ret

 End of assembler dump.

3) foo3.gdb.out

 (gdb) disas /m main
 Dump of assembler code for function main:
        {
    0x080483ed <+>:     push   ebp
    0x080483ee <+>:     mov    ebp,esp
    0x080483f0 <+>:     sub    esp,0x20

                unsigned int a[] = {, , };
    0x080483f3 <+>:     mov    DWORD PTR [ebp-0xc],0x1
    0x080483fa <+>:    mov    DWORD PTR [ebp-0x8],0x2
    0x08048401 <+>:    mov    DWORD PTR [ebp-0x4],0x3

                unsigned int n = sizeof (a) / sizeof (int);
    0x08048408 <+>:    mov    DWORD PTR [ebp-0x10],0x3

                unsigned int sum = ;
    0x0804840f <+>:    mov    DWORD PTR [ebp-0x18],0x0

                for (unsigned int i = ; i < n; i++)
    0x08048416 <+>:    mov    DWORD PTR [ebp-0x14],0x0
    0x0804841d <+>:    jmp    0x8048437 <main+>
    0x08048433 <+>:    add    DWORD PTR [ebp-0x14],0x1
    0x08048437 <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x0804843a <+>:    cmp    eax,DWORD PTR [ebp-0x10]
    0x0804843d <+>:    jb     0x804841f <main+>

                        sum += *(i+a);
    0x0804841f <+>:    mov    eax,DWORD PTR [ebp-0x14]
    0x08048422 <+>:    lea    edx,[eax*+0x0]
    0x08048429 <+>:    lea    eax,[ebp-0xc]
    0x0804842c <+>:    add    eax,edx
    0x0804842e <+>:    mov    eax,DWORD PTR [eax]
    0x08048430 <+>:    add    DWORD PTR [ebp-0x18],eax

               return sum;
    0x0804843f <+>:    mov    eax,DWORD PTR [ebp-0x18]

       }
    0x08048442 <+>:    leave
    0x08048443 <+>:    ret

 End of assembler dump.

4) a[i] v.s. i[a]

5) i[a] v.s. *(i+a)

结论： a[i]==i[a]==*(i+a)==*(a+i)

分析： 在编译器的眼里，数组名a不过是一段连续内存的首地址。获取某个元素a[i]不过是在a对应的首地址上做偏移，找到对应的内存地址后从中取出其中的内容即可。

PS： 我在面试别人的过程中，如果求职的工程师说他懂汇编，我一般会问这样的问题，"能否在C代码中使用i[a]去访问数组a的第i个元素？"。无论对方说能与不能，我都很乐意进一步问"为什么能/不能?"从而挖掘出其对汇编及编译过程的理解深度。 通常，优秀的程序员能回答得富有计算机思维（即使他判定为不能， 比如"我觉得不能，编译器应该不支持这种怪诞的用法..."），而那些机械的程序员一般会选择放弃思考为什么能/不能。