c - 汇编代码和 Switch 语句案例

Question

我无法理解如何通过汇编语言确定案例 50、52 等。

据我了解，跳转表对应于每种情况下要执行的操作，并且检查 edx > 5 意味着情况范围从 0 到 5？我相信 1 被遗漏了，因为它是默认情况，但为什么遗漏了 5？

我感觉应该有一个case 55: where result *= result, no?

如果有人能帮忙解释一下，那就太好了。谢谢!

int switch_prob(int x) 
{
    int result = x;
    switch (x)
    {
        case 50:
        case 52:
            result <<= 2;
            break;
        case 53:
            result >>= 2;
            break;
        case 54:
            result *= 3;
            break;
        default:
            result += 10;   
    }
    return result;
}

Figure 3.38 shows the disassembled object code for the procedure. We are only interested in the part of the code shown on lines 4 through 16. We can see on line 4 that parameter x (at offset 8 relative to %ebp) is loaded into register %eax, corresponding to program variable result. The “lea 0x0(%esi), %esi” instruction on line 11 is a nop instruction inserted to make the instruction on line 12 start on an address that is a multiple of 16

The jump table resides in a different area of memory. Using the debugger GDB, we can examine the six 4-byte words of memory starting at address 0x8048468 with the command x/6w 0x8048468. GDB prints the following:

(gdb) x/6w
0x8048468: 0x080483d5 0x080483eb 0x080483d5 0x0x80483e0
0x8048478: 0x080483e5 0x080483e8
(gdb)

Assembly Code:

1: 080483c0 <switch_prob>:
2: 80483c0: push %ebp
3: 80483c1: mov %esp,%ebp
4: 80483c3: mov 0x8(%ebp),%eax // X is copied into eax ; eax = x
5: 80483c6: lea 0xffffffce(%eax),%edx // placeholder 
6: 80483c9: cmp $0x5, %edx // Compare edx (3rd argument) with 5; Edx - 5 // clearly, edx is x
7: 80483cc: ja 80483eb <switch_prob+0x2b> // if edx - 5 > 0, Jump into line 16 (default)
8: 80483ce: jmp *0x8048468(,%edx,4) // Go into the jump table
9: 80483d5: shl $0x2, %eax // eax << 2
10: 80483d8: jmp 80483ee <switch_prob+0x2e> // Jump to line 17
11: 80483da: lea 0x0(%esi),%esi // esi = esi  NOP. Filling in N bytes
12: 80483e0: sar $0x2, %eax // eax >> 2
13: 80483e3: jmp 80483ee <switch_prob+0x2e> // Jump to line 17
14: 80483e5: lea (%eax, %eax, 2), %eax // eax = eax + 2(eax)
15: 80483e8: imul %eax, %eax // eax *= eax
16: 80483eb: add $0xa, %eax // eax += 10
17: 80483ee: mov %ebp, %esp // esp = ebp
18: 80483f0: pop %ebp
19: 80483f1: ret

Answer 1

程序集与源代码不匹配。它匹配更像这样的东西:

int switch_prob(int x) 
{
    int result = x;
    switch (x)
    {
        case 50:
        case 52:
            result <<= 2;
            break;
        case 53:
            result >>= 2;
            break;
        case 54:
            result *= 3;
            // WARNING: Falls through
        case 55:
            result *= result;
            // WARNING: Falls through
        default:
            result += 10;   
            break;
    }
    return result;
}

这很可能是人为错误造成的(例如，更新问题中的源代码，使其与去年学生得到的问题不相同，但忘记更新程序集以匹配)。

永远不要假设教师/教授不是人类......