在汇编语言寻址模式（IA-32 NASM） [英] Addressing Modes in Assembly Language (IA-32 NASM)

问题描述

由于在这个网上资源是稀少的，我会为未来搜索的好处，首先列出了IA-32汇编语言的寻址模式（NASM），然后跟进一个简单的问题。

1. 寄存器寻址
   
   
   
   • MOV EAX，EBX：复制什么是EBX到EAX
   
   
   • MOV ESI，VAR：复制VAR的地址（比如0x0040120e）进入ESI
   
   


2. 立即寻址（第二个操作数是一个立即数）
   
   
   
   • MOV BX，20：16位寄存器BX得到实际值20
   
   


3. 直接内存寻址（通过指定地址从内存中直接加载）
   
   
   
   • MOV AX，[1000H]：在地址4096点（0x1000十六进制）加载从字节2个字节的对象到一个名为AX
   16位寄存器
   
   • MOV [1000H]，斧：内存地址1000H得到斧头的价值
   
   


4. 直接偏移量寻址（同3，只是用算术来修改地址）
   
   
   
   • MOV人，[byte_tbl + 2]
   
   


5. （通过使用存储在寄存器地址访问内存）寄存器间接
   
   
   
   • MOV AX，[二]：在迪指定的内存地址拷贝值，到AX
   
   
   • MOV DWORD [EAX]，VAR1：在VAR1值复制到由EAX指定内存插槽
   
   

请注意，上述为NASM。对于MASM / TASM你会使用MOV ESI，OFFSET富来获得地址，而MOV ESI，foo和MOV ESI，[富]都将获得价值（creds到@迈克尔）。

所以，到我的问题。这是在下面的教程29页的底部相对于一个例子：<一href=\"http://www.tutorialspoint.com/assembly_programming/assembly_tutorial.pdf\">http://www.tutorialspoint.com/assembly_programming/assembly_tutorial.pdf

基本上，它列出了下面code作为间接内存寻址的例子。

  MY_TABLE次，每次10 DW 0;分配10个字（2个字节），每个初始化为0
MOV EBX，[MY_TABLE]在EBX MY_TABLE的有效地址
MOV [EBX]，110; MY_TABLE [0] = 110
ADD EBX，2; EBX = EBX +2
MOV [EBX]，123; MY_TABLE [1] = 123
 

我的问题：

1. 应该不是MOV EBX，[MY_TABLE]其实是MOV EBX，MY_TABLE，因为我们希望把EBX表的地址，而不是本身的价值？


2. 肯定是MY_TABLE [2]，等于123结尾，而不是MY_TABLE [1]？

解决方案

1. 在NASM语法，该指令应 MOV EBX，MY_TABLE 。什么 MOV EBX，[MY_TABLE] 会做的是位于 MY_TABLE 前4个字节加载到 EBX 。另一种方法是使用 LEA ，如 LEA EBX，[MY_TABLE] 。




2. 在这种情况下，本教程是正确的。 MY_TABLE 被定义为词的数组。在x86一个字为2个字节，因此 MY_TABLE 确实位于第二个元素MY_TABLE + 2 。

As the web-resources on this is sparse, I will, for the benefit of future searches, begin by listing the address modes for IA-32 Assembly Language (NASM) and then follow up with a quick question.

1. Register addressing
   • mov eax, ebx: Copies what is in ebx into eax
   • mov esi, var: Copies address of var (say 0x0040120e) into esi
2. Immediate addressing (second operand is an immediate constant)
   • mov bx, 20: 16-bit register bx gets the actual value 20
3. Direct memory addressing (directly loads from memory through a specified address)
   • mov ax, [1000h]: loads a 2-byte object from the byte at address 4096 (0x1000 in hexadecimal) into a 16-bit register called 'ax'
   • mov [1000h], ax: memory at address 1000h gets the value of ax
4. Direct offset addressing (same as 3, just using arithmetics to modify address)
   • mov al, [byte_tbl+2]
5. Register indirect (accessing memory by using addresses stored in registers)
   • mov ax, [di]: copies value at memory address specified by di, into ax
   • mov dword [eax], var1: copies value in var1 into the memory slot specified by eax

Please note that the above is for NASM. For MASM/TASM you'd use "mov esi, OFFSET foo" to get the address, while "mov esi, foo" and "mov esi, [foo]" both would get the value (creds to @Michael).

So, onto my question. It is in in relation to an example at the bottom of page 29 of the following tutorial: http://www.tutorialspoint.com/assembly_programming/assembly_tutorial.pdf

It basically lists the below code as an example of indirect memory addressing.

MY_TABLE TIMES 10 DW 0 ; Allocates 10 words (2 bytes) each initialized to 0 
MOV EBX, [MY_TABLE] ; Effective Address of MY_TABLE in EBX 
MOV [EBX], 110 ; MY_TABLE[0] = 110 
ADD EBX, 2 ; EBX = EBX +2 
MOV [EBX], 123 ; MY_TABLE[1] = 123 

My questions:

1. Should not "MOV EBX, [MY_TABLE]" in fact be "MOV EBX, MY_TABLE", as we want to put the address of the table in EBX, not the value itself?
2. Surely it is MY_TABLE[2] that is equal to 123 at the end, not MY_TABLE[1]?

解决方案

1. In NASM syntax, that instruction should be MOV EBX, MY_TABLE. What MOV EBX, [MY_TABLE] would do is load the first 4 bytes located at MY_TABLE into EBX. Another alternative would be to use LEA, as in LEA EBX, [MY_TABLE].

2. In this case the tutorial is right. MY_TABLE is defined as an array of words. A word on the x86 is 2 bytes, so the second element of MY_TABLE is indeed located at MY_TABLE + 2.

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