大会（EMU8086）不允许字节移入8位寄存器 [英] Assembly (emu8086) not allowing moving of bytes into 8-bit registers

问题描述

我想提出一个计算器类型的节目，我用它来从用户那里得到一个数字，并将其存储：

I am making a calculator-type program, and I use this to get a number from the user and store it:

mov ah, 01h
int 21h
mov offset num1, al

和在code结束时，我有NUM1设置为一个字节以

and at the end of the code I have num1 set up as a byte with

num1 db 0

给它的默认值为0。

giving it a default value of 0.

问题是，当我尝试从NUM1值搬回到寄存器来执行实际操作：

The problem is, when I try to move the value from num1 back into a register to perform the actual operation:

mov bl, offset num1

我得到一个错误说第二个操作数是超过8位，我无法通过互联网/手册中的任何搜索摸不着头脑。

I get an error saying the second operand is over 8 bits, and I cannot figure this out through any searching of the internet/manuals.

另外，我使用偏移瓦尔，因为这就是我最初是教他们，我真的不明白他们的任何其他方式。

Also, I use offset vars because that is how I was initially taught them and I don't really understand them any other way.

推荐答案

这

mov offset num1, al

应该是：

mov num1, al

和

mov bl, offset num1

应该是：

mov bl, num1

的偏移关键字应该在你想要得到一个标签的地址（或它的一个段内的偏移，以precise）的情况下使用。例如，如果你使用 INT 21H / AH = 09H 来打印你需要一个字符串 DX 持有的偏移字符串打印，所以你可以使用 MOV DX，偏移my_string 。结果
但在code你已经证明你在 NUM1 加载和存储的价值有兴趣，所以你只需要使用 NUM1 （或 [NUM1] ，这意味着为 NUM1 在MASM / TASM同样的事情在这种情况下的语法）。

The offset keyword should be used in situations where you want to get the address of a label (or its offset within a segment, to be precise). For example, if you use INT 21H / AH=09H to print a string you need DX to hold the offset of the string to print, so you would use mov dx, offset my_string.
But in the code you've shown you're just interested in loading and storing the value at num1, so you should simply use num1 (or [num1], which means the same thing as num1 in MASM/TASM syntax in this context).

对于错误信息的含义是：一个在真实模式程序偏移量是16位的，所以即使你真的想将它移动到一个8位寄存器，如人或 BL ，你就不能这样做。

Regarding the meaning of the error message: an offset in a real-mode program is 16 bits, so even if you really wanted to move it into an 8-bit register like al or bl you wouldn't be able to do so.

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