CSAPP 习题3.36

题目如下

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You are charged with maintaining a large C program, and you come across the following code:

 typedef struct {
   int left;
   a_struct a[CNT];
   int right;
 } b_struct;

 void test(int i, b_struct *bp)
 {
   int n = bp->left + bp->right;
   a_struct *ap = &bp->a[i];
   ap->x[ap->idx] = n;
 }

Unfortunately, the ‘.h’ file defining the compile-time constant CNT and the structure a_struct are in files for which you do not have access privileges. Fortunately, you have access to a ‘.o’ version of code, which you are able to disassemble with the objdump program, yielding the disassembly shown in Figure below.

 00000000 <test>:
 0: 55 push %ebp
 1: 89 e5 mov %esp,%ebp
 3: 53 push %ebx
 4: 8b 45 08 mov 0x8(%ebp),%eax
 7: 8b 4d 0c mov 0xc(%ebp),%ecx
 a: 8d 04 80 lea (%eax,%eax,4),%eax
 d: 8d 44 81 04 lea 0x4(%ecx,%eax,4),%eax
 11: 8b 10 mov (%eax),%edx
 13: c1 e2 02 shl $0x2,%edx
 16: 8b 99 b8 00 00 00 mov 0xb8(%ecx),%ebx
 1c: 03 19 add (%ecx),%ebx
 1e: 89 5c 02 04 mov %ebx,0x4(%edx,%eax,1)
 22: 5b pop %ebx
 23: 89 ec mov %ebp,%esp
 25: 5d pop %ebp
 26: c3 ret

Using your reverse engineering skills, deduce the following:

A. The value of CNT.

B. A complete declaration of structure a_struct. Assume that the only fields in this structure are idx and x.

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将汇编代码按句翻译成伪代码

%eax = i

%ecx = bp

%eax = 5i

%eax = bp + 20i + 4                      // int left占4 每个a_struct占20

%eax = &(bp->a[i])

%edx = *(int*) &(bp->a[i])               // int idx

%edx *= 4                                // idx*4

%ebx = bp + (4 + 20*9)                   // right所在位置，所以CNT = 9

%ebx = left + right

%ebx = &(bp->a[i]) + idx*4 + 4           // 所以 a_struct是一个int加一个4 int的数组。

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记录下每一小步，就当备忘吧