在x86汇编混淆添加命令 [英] Confusing add command in x86 assembly

问题描述

我一直在寻找通过一些code和发现2号线是困扰我：

I was looking through some code and found 2 lines that perplexed me:

add    -0x4(%esi,%ebx,4),%eax
cmp    %eax,(%esi,%ebx,4)

我习惯了标准的添加SRC，DST 和 CMP X1，X2 和我不是真的知道什么这些线路实际上做的。

I am accustomed to the standard add src,dst and cmp x1,x2 and I'm not really sure what these lines are actually doing.

我相信这是使用GCC编译

I believe that it is compiled with GCC

推荐答案

这是一个使用基地+（索引*比例）+位移寻址模式。至少，我是这么认为的。我不是真正熟悉AT＆amp; T公司的语法。我认为，英特尔的语法是：

That's using the Base + (Index * Scale) + Displacement addressing mode. At least, I think so. I'm not real familiar with the AT&T syntax. I think the Intel syntax would be:

add eax,[esi + ebx*4 - 4]
cmp [esi + ebx*4],eax

这看起来像它的索引为整数（4字节的值）的数组。用C试想一下，你想从一些数组元素的值与总，是这样的：

This looks like it's indexing into an array of integers (4-byte values). Imagine in C that you want to add the value from some array element to a total, like this:

int a[100];
int i = 10;
int total = 0;
total += a[i-1];

现在，让 ESI 持有数组的地址， EBX 持有我和 EAX 持有的价值33你会得到：

Now, make esi hold the address of the array, ebx hold the value of i, and eax hold the value 33. You'd get:

add eax,[esi + ebx*4 - 4]

的比较指令被测试，看看是否结果（ eax中）等于所述阵列中的下一个值。在C的例子，这将是等同于总比较 A [I] 。

The comparison instruction is testing to see if the result (in eax) is equal to the next value in the array. In the C example, that would be equivalent to comparing total to a[i].

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