x86指令含义 [英] x86 instruction meaning

问题描述

我通过一些code运行现在的gdb的，我不知道什么这两个指令实际上做。如果有人可以帮助我，我真的AP preciate它。

 添加-0x2c（EBP％，EBX％，4），EAX％
CMP％EAX，-0x28（EBP％，EBX％，4）
 

解决方案

x86汇编通常更容易，当你在Intel语法，而不是AT＆放写明白; T语法

在Intel语句将是：

 添加EAX，[EBP + 4 * EBX-0x2C]
CMP [EBP + 4 * EBX-0x28]，EAX
 

第一个指令（添加）将存储在存储器地址的值 [EBP + 4 * EBX-0x2C] 为 EAX ，并将之和 EAX 。

第二个指令（ CMP ）比较 EAX 与 [EBP + 4 * EBX -0x28] 从存储在内存中的地址的字值减去 EAX 的值 [EBP + 4 * EBX-0x28] ，并设置标志（的， SF ， ZF ， AF ， PF ， CF ），据此，但不会在任何地方保存结果。 CMP 是完全一样的分，唯一的区别是一个事实，即在子的结果保存在 CMP 不是

比较的类型，通常的形式创建的一个条件跳转。在x86汇编有很多条件跳转，他们分支是否依赖于标志的值。

I'm running through some code right now on gdb and I have no clue what these two instructions actually do. If anyone could help me out, I'd really appreciate it.

add  -0x2c(%ebp, %ebx, 4), %eax
cmp  %eax, -0x28(%ebp, %ebx, 4)

解决方案

x86 assembly is usually much easier to understand when you write it in Intel syntax instead of AT&T syntax.

In Intel syntax it would be:

add eax,[ebp+4*ebx-0x2C]
cmp [ebp+4*ebx-0x28],eax

The first instruction (add) adds the value of word stored in the memory address [ebp+4*ebx-0x2C] to the value of eax, and stores the sum in eax.

The second instruction (cmp) compares eax with [ebp+4*ebx-0x28] by subtracting the value of eax from the value of the word stored in the memory address [ebp+4*ebx-0x28], and sets flags (OF, SF, ZF, AF, PF, CF) accordingly but does not save the result anywhere. cmp is exactly the same as sub, the only difference being the fact that in sub the result is saved, in cmp not.

The type of comparison is usually created in the form a conditional jump. In x86 assembly there are a lot of conditional jumps and whether they branch depends on the values of the flags.

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