基本使用的x86汇编和YASM立即数（方括号）的 [英] Basic use of immediates (square brackets) in x86 Assembly and yasm

问题描述

假如我有以下声明：

section .bss
buffer    resb     1

和这些说明如下：

mov    al, 5
mov    [buffer], al
mov    bl, [buffer]
mov    cl, buffer

我是正确的理解是BL将包含值5，和CL将包含变量的内存地址缓存？

我感到困惑

• 移动立即到寄存器，


• 移动寄存器到直接（在发生什么，数据或地址？）和


• 移动立即到寄存器没有括号
  
  
  
  • 例如， MOV CL，缓冲 VS MOV CL [缓冲]
  
  
  • moving an immediate into a register,
  • moving a register into an immediate (what goes in, the data or the address?) and
  • moving an immediate into a register without the brackets
    • For example, mov cl, buffer vs mov cl, [buffer]

    更新：看完后的反应，我想下面的总结准确的：

    UPDATE: After reading the responses, I suppose the following summary is accurate:

    假设声明阵列RESB 0 在存在.bss段。我的理解是：

    Assume the declaration array resb 0 exists under section .bss. My understanding is that:

    
    
    • MOV EDI，阵列会将零数组索引的内存地址 EDI 。
    
    
    • MOV [EDI]，3 会将值3到数组
    的零指数
    
    • 在添加EDI，3 ， EDI 现在包含数组的第3个指数的内存地址
    
    
    • MOV人，[阵列] 把该数据的零指数为人。
    
    
    • MOV人，[阵列+ 3] 放置数据的第三个指数为人。
    
    
    • MOV [人]，[阵列] 是无效的，无论出于何种原因。
    
    
    • MOV阵列，3 是无效的，因为你不能说嘿，我不喜欢的偏移量，阵列存储，所以我会叫它3
    
    
    • MOV [阵列] 3 把值3到数组的第零指标。
    
    
    • mov edi, array puts the memory address of the zeroth array index in edi.
    • mov [edi], 3 puts the VALUE 3 into the zeroth index of the array
    • after add edi, 3, edi now contains the memory address of the 3rd index of the array
    • mov al, [array] puts the DATA at the zeroth index into al.
    • mov al, [array+3] puts the DATA at the third index into al.
    • mov [al], [array] is invalid, for whatever reason.
    • mov array, 3 is invalid, because you can't say "Hey, I don't like the offset at which array is stored, so I'll call it 3"
    • mov [array], 3 puts the value 3 into the zeroth index of the array.

    请注明如果这些都是假的。

    Please mention if any of these is false.

    推荐答案

    事实上，你的思想是correct.That是，BL将包含5和CL缓冲器的存储器地址（实际上标签缓冲器是存储器地址本身）

    Indeed, your thought is correct.That is, bl will contain 5 and cl the memory address of buffer(in fact the label buffer is a memory address itself).

    现在，让我解释一下你提到的操作之间的差异：

    Now, let me explain the differences between the operations you mentioned:

    
    

    • 移动立即到是可以做到用一个寄存器 MOV章，IMM 。什么可能是令人困惑的是，标签如缓冲区是包含即时数值本身地址。

    • moving an immediate into a register can be done using mov reg,imm.What may be confusing is that labels e.g buffer are immediate values themselves that contain an address.

    您不能真正移动寄存器到直接，因为眼前的值是常量，如 2 或 FF1AH 。什么，你可以做的是移动寄存器其中常数点to.You可以像 MOV [常量]，章。

    You cannot really move a register into an immediate, since immediate values are constants, like 2 or FF1Ah.What you can do is move a register to the place where the constant points to.You can do it like mov [const], reg .

    您也可以使用间接寻址像 MOV REG 2，[REG1] 提供REG1指向有效的位置，而且将转移由REG1指向的值REG2。

    You can also use indirect addressing like mov reg2,[reg1] provided reg1 points to a valid location, and it will transfer the value pointed by reg1 to reg2.

    因此​​， MOV CL，缓冲将移动缓冲区的地址以CL（这可能会或可能不会给出正确的地址，因为CL只有一个字节长），而 MOV CL [缓冲] 将得到的实际值。

    So, mov cl, buffer will move the address of buffer to cl(which may or may not give the correct address, since cl is only one byte long) , whereas mov cl, [buffer] will get the actual value.

    
    
    • 当你用[A]，那么你参考价值的地方，一个点to.For例如，如果是 F5B1 ，然后按[A]是指在地址F5B1的内存的
    
    
    • 标签的地址，即价值如 F5B1 。
    
    存储在寄存器
    • 值不必被引用为[章]，因为寄存器不必addresses.In事实上，寄存器可以被认为是即时值。
    
    
    • When you use [a], then you refer to the value at the place where a points to.For example, if a is F5B1, then [a] refers to the address F5B1 in RAM.
    • Labels are addresses,i.e values like F5B1.
    • Values stored in registers do not have to be referenced to as [reg] because registers do not have addresses.In fact, registers can be thought of as immediate values.

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