ADDL指令，86 [英] Addl instruction, x86

问题描述

您必须原谅我，我一般品牌的新x86汇编，和组装。

You'll have to excuse me, I'm brand new to x86 assembly, and assembly in general.

所以我的问题是，我有这样的：

So my question is, I have something like:

addl %edx,(%eax)

％eax中是持有一个指向某个整数寄存器。让XP把它叫做

%eax is a register which holds a pointer to some integer. Let's call it xp

这是否意味着它的说法： * XP = * XP +％EDX ？ （％EDX 是一个整数）

Does this mean that it's saying: *xp = *xp + %edx? (%edx is an integer)

我只是困惑在哪里ADDL将存储结果。如果％EAX 是一个指向一个int，那么（％EAX）应该是INT的实际值。因此，将 ADDL 存储％EDX +（％EAX）的结果在 * XP ？我更喜欢有人给我讲解一下！

I'm just confused where addl will store the result. If %eax is a pointer to an int, then (%eax) should be the actual value of that int. So would addl store the result of %edx+(%eax) in *xp? I would really love for someone to explain this to me!

我真的AP preciate任何帮助！

I really appreciate any help!

推荐答案

是的，这个指令做什么，你觉得它在做什么。

Yes, this instruction is doing exactly what you think it's doing.

大多数x86算术指令有两个操作数：源和目标。在AT＆amp; T公司的语法（这里使用），目的地是永远正确的操作。因此一个指令，如：

Most x86 arithmetic instructions take two operands: a source and a destination. In AT&T syntax (used here), the destination is always the right operand. So with an instruction like:

addl %edx, %eax

中的值 EDX 和 EAX 相加，结果存储在 EAX 。然而，在你的榜样，内存间接寻址使用（这是由括号表示）。这意味着 EAX 作为指针处理，所以正确的操作数取自地址由 EAX 指出，并将其结果存储到同一地址。

the values in edx and eax are added together and the result is stored in eax. However, in your example, memory indirect addressing is used (this is denoted by parentheses). This means that eax is treated as a pointer, so the right operand is taken from the address pointed to by eax, and the result is stored to the same address.

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