YASM/NASM x86组件中立即数与方括号的基本用法 [英] Basic use of immediates vs. square brackets in YASM/NASM x86 assembly

问题描述

假设我声明了以下内容:

Suppose I have the following declared:

section .bss
buffer    resb     1

这些说明位于section .text中:

mov    al, 5                    ; mov-immediate
mov    [buffer], al             ; store
mov    bl, [buffer]             ; load
mov    cl, buffer               ; mov-immediate?

我正确理解bl将包含值5，而cl将包含变量buffer的内存地址吗?

Am I correct in understanding that bl will contain the value 5, and cl will contain the memory address of the variable buffer?

我对

• 将立即数移到寄存器中，
• 将寄存器移至立即数(输入的是数据还是地址?)和
• 将立即数移到没有括号的寄存器中
  • 例如mov cl, buffer vs mov cl, [buffer]
  • 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:阅读回复后，我认为以下摘要正确无误:

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

    • mov edi, array将第零个数组索引的内存地址放入edi.即标签地址.
    • mov byte [edi], 3将值3放入数组的第零个索引
    • add edi, 3之后，edi现在包含数组第三个索引的内存地址
    • mov al, [array]将第零个索引处的DATA装入al.
    • mov al, [array+3]将第三个索引处的DATA装入al.
    • mov [al], [array]无效，因为 x86无法对2个显式编码内存操作数，并且由于al只有8位，即使在16位寻址模式下也无法使用. 引用内存位置的内容. (x86寻址模式)
    • mov array, 3无效，因为您不能说嘿，我不喜欢存储array的偏移量，所以我将其称为3".立即数只能是源操作数.
    • mov byte [array], 3将值3放入数组的第零个索引(第一个字节). 需要byte说明符，以避免字节之间的歧义/word/dword包含内存和立即操作数的指令.否则，这将是一个汇编时错误(操作数大小不明确).
    • mov edi, array puts the memory address of the zeroth array index in edi. i.e. the label address.
    • mov byte [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] loads the DATA at the zeroth index into al.
    • mov al, [array+3] loads the DATA at the third index into al.
    • mov [al], [array] is invalid because x86 can't encode 2 explicit memory operands, and because al is only 8 bits and can't be used even in a 16-bit addressing mode. Referencing the contents of a memory location. (x86 addressing modes)
    • 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". An immediate can only be a source operand.
    • mov byte [array], 3 puts the value 3 into the zeroth index (first byte) of the array. The byte specifier is needed to avoid ambiguity between byte/word/dword for instructions with memory, immediate operands. That would be an assemble-time error (ambiguous operand size) otherwise.

    如果其中任何一项为假，请提及. (编者注:我修复了语法错误/歧义，因此有效的错误实际上是有效的NASM语法.并链接了其他Q&A以获得详细信息)

    Please mention if any of these is false. (editor's note: I fixed syntax errors / ambiguities so the valid ones actually are valid NASM syntax. And linked other Q&As for details)

    推荐答案

    实际上，您的想法是正确的.也就是说，bl将包含5和cl的缓冲区内存地址(实际上，标签缓冲区本身就是内存地址)

    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，让我解释一下您提到的操作之间的区别:

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

    • 将立即数移动到寄存器中可以使用mov reg,imm完成.可能令人困惑的是，例如缓冲区之类的标签本身就是包含地址的立即数.

    • 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.

    您不能真正将寄存器移到立即数中，因为立即数是常量，例如2或FF1Ah.您可以做的就是将寄存器移到常量所指向的位置.像mov [const], reg一样.

    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 .

    还可以使用mov reg2,[reg1]之类的间接寻址，只要reg1指向有效位置，它将reg1指向的值传输到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, buffer会将缓冲区的地址移动到cl(由于cl只有一个字节长，它可能会或可能不会提供正确的地址)，而mov cl, [buffer]会得到实际值.

    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.

    • 使用[a]时，指的是指向某处的值.例如，如果a为F5B1，则[a]指的是 RAM中的地址F5B1 .
    • 标签是地址，即F5B1之类的值.
    • 因为寄存器没有地址，所以不必将存储在寄存器中的值称为[reg].实际上，可以将寄存器视为立即值.
    • 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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