问题描述
您必须原谅我，我是x86组装和一般组装的新手。
You'll have to excuse me, I'm brand new to x86 assembly, and assembly in general.
所以我的问题是，我有一些类似的东西：
So my question is, I have something like:
addl %edx,(%eax)
％eax是一个保存指向某个整数的指针的寄存器。我们称它为xp
%eax is a register which holds a pointer to some integer. Let's call it xp
这是否表示： * xp = * xp +％edx ？ （％edx 是整数）
Does this mean that it's saying: *xp = *xp + %edx? (%edx is an integer)
我只是迷惑在addl存储结果的位置。如果％eax 是一个指向int的指针，则（％eax）应该是该int的实际值。因此， addl 会将％edx +（％eax）的结果存储在 * xp ？我真的很希望有人向我解释一下！
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!
我非常感谢您的帮助！
推荐答案
是的，此指令完全按照您的想法做。
Yes, this instruction is doing exactly what you think it's doing.
大多数x86算术指令采用两个操作数：源和目标。在AT& T语法（在此使用）中，目的地总是正确的操作数。因此，使用以下指令：
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
edx 和 eax 中的值相加并将结果存储在 eax 中。但是，在您的示例中，（％eax）是一个内存操作数；这就是AT& T语法中括号的含义（就像NASM语法中的方括号一样）。
the values in edx and eax are added together and the result is stored in eax. However, in your example, (%eax) is a memory operand; that's what parentheses mean in AT&T syntax (like square-brackets in NASM syntax).
这意味着 eax 被视为指针，因此从 eax 指向的地址中获取正确的操作数，并将结果存储到相同的地址。
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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