汇编器mov问题
我有下一个代码：
mov ax,@data
mov ds,ax
为什么我不能这样写？
mov ds,@data
全部来源：
.MODEL small
.STACK 100h
.DATA
HelloMessage DB 'Hello, world',13,10,'$'
.CODE
.startup
mov ax,@data
mov ds,ax
mov ah,9
mov dx,OFFSET HelloMessage
int 21h
mov ah,4ch
int 21h
END
谢谢！
I have the next code:
mov ax,@data
mov ds,ax
Why I can not write just like this?
mov ds,@data
All source:
.MODEL small
.STACK 100h
.DATA
HelloMessage DB 'Hello, world',13,10,'
Thank you!
.CODE
.startup
mov ax,@data
mov ds,ax
mov ah,9
mov dx,OFFSET HelloMessage
int 21h
mov ah,4ch
int 21h
END
Thank you!
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你不能，因为指令集不包含执行此操作的指令。这只是 x86 的众多特性之一。
对于汇编语言来说，这些限制是相当正常的。大多数架构都包含一些经过特殊处理的寄存器（例如处理器状态字），但通常少于 x86 架构。
不为所有可能的移动提供指令的原因是为了减少指令集的大小，以便指令占用更少的内存。总体而言，分两步完成很少需要的动作会更有效。
You can't, because the instruction set doesn't contain an instruction to do that. It is just one of the many idiosyncrasies of the x86.
These kind of restrictions are fairly normal for assembly languages. Most architectures contain some registers that are treated specially (for example the processor status word), though usually fewer than the x86 architecture.
The reason to not provide an instruction for all possible moves is to reduce the size of the instruction set, so that an instruction takes less memory. Overall it is more efficient to do moves that are rarely needed in two steps.
通用寄存器作为“ax”被设计为保存指向数据的16位数字（在您的情况下是DATA内部的字符串）
所以如果您尝试直接将数据传递到特殊寄存器（此处为ds或数据段）它不会正常工作，因为它不知道以这种方式接受数据。因此，我们首先获取该“数字”或数据开始的内存位置点。将该点传递给 ds 寄存器。
General purpose register as the 'ax' is designed to hold the 16 bit number pointing to the data (in your case the string inside the DATA)
So if you try to directly pass the data to the special register (ds or data segment here) it will not work correctly as it does not know to accept data that way. So we first get that 'number' or the point in memory location where data starts & pass that point to ds register.
我不是专家，但这就是我对这种限制的理解。
段寄存器用于控制寄存器指令使用哪个内存段，因此您最不想做的就是从内存位置加载段寄存器（在本例中为 DS，即数据段寄存器）。修改 DS 的行为可能会导致正在读取的内存位置在更新 DS 的过程中发生变化，即加载到 DS 中的第一个位/字节现在导致它在读取其余部分之前指向另一个段。将值读入累加器 (AX) 或另一个通用寄存器会更安全，因此现在该值在加载到段寄存器时位于处理器中，因此在加载过程中该值不会被损坏。
I'm no expert but this is how I understand this constraint to work.
The Segment registers are used to control which segment of memory is used by the register instructions, as such the last thing you want is to load a segment register (DS in this case which is the Data Segment register) from a memory location. The act of modifying DS could cause the memory location being read to change during the process of updating DS, i.e. the first bits/byte loaded into DS now cause it to be pointing to another segment before the remainder has been read. It's safer to read the value into the Accumulator (AX) or another general purpose register, so now the value is in the processor when it's loaded into the segment register, so no chance of the value getting corrupted during the load.