汇编语言中的寻址模式 (IA-32 NASM)

那么，进入我的问题.它与以下教程第 29 页底部的示例有关:http://www.tutorialspoint.com/assembly_programming/assembly_tutorial.pdf它基本上列出了以下代码作为间接内存寻址的示例.MY_TABLE TIMES 10 DW 0 ;分配 10 个字(2 个字节)，每个字都初始化为 0MOV EBX, [MY_TABLE] ;EBX 中 MY_TABLE 的有效地址MOV [EBX], 110 ;MY_TABLE[0] = 110添加 EBX, 2 ;EBX = EBX +2MOV [EBX], 123 ;MY_TABLE[1] = 123我的问题:MOV EBX, [MY_TABLE]"实际上不应该是MOV EBX, MY_TABLE"，因为我们想把表的地址放在EBX中，而不是值本身?肯定是 MY_TABLE[2] 最后等于 123，而不是 MY_TABLE[1]? 解决方案 在 NASM 语法中，该指令应该是 MOV EBX, MY_TABLE.MOV EBX, [MY_TABLE] 要做的是将位于 MY_TABLE 的前 4 个字节加载到 EBX 中.另一种选择是使用 LEA，如LEA EBX，[MY_TABLE].在这种情况下，教程是正确的.MY_TABLE 被定义为一个单词数组.x86上的一个字是2个字节，所以MY_TABLE的第二个元素确实位于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.Register addressingmov eax, ebx: Copies what is in ebx into eaxmov esi, var: Copies address of var (say 0x0040120e) into esiImmediate addressing (second operand is an immediate constant)mov bx, 20: 16-bit register bx gets the actual value 20Direct 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 axDirect offset addressing (same as 3, just using arithmetics to modify address)mov al, [byte_tbl+2]Register indirect (accessing memory by using addresses stored in registers)mov ax, [di]: copies value at memory address specified by di, into axmov dword [eax], var1: copies value in var1 into the memory slot specified by eaxPlease 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.pdfIt 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 0MOV EBX, [MY_TABLE] ; Effective Address of MY_TABLE in EBXMOV [EBX], 110 ; MY_TABLE[0] = 110ADD EBX, 2 ; EBX = EBX +2MOV [EBX], 123 ; MY_TABLE[1] = 123My questions: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?Surely it is MY_TABLE[2] that is equal to 123 at the end, not MY_TABLE[1]? 解决方案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].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. 这篇关于汇编语言中的寻址模式 (IA-32 NASM)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！