中国DOS联盟

I believe many people learn DOS assembly by reading this book?
When introducing addressing modes inside, it explains that the sentence mov ax, is: assign the content at ds: to ax, that is, if the value at ds: is 3020, then after executing this sentence, ax = 3020H.
Recently, when debugging a DOS program, I found that instructions in the form of mov xx,, add xx, (where there are all numbers) are all interpreted by the assembler as mov xx,xxxxh, that is, directly give the value inside the square brackets to the register! That is, after mov ax, is executed, ax = 2000H. Don't believe it? You can compile a few sentences yourself, and then load them into the debugger to check the code in memory, and you will find that after executing this sentence, ax = 2000h.
Unless there is a problem with my machine or the programming software (masm6.11, tasm5, Tdebug). Otherwise, please tell me how to understand this addressing mode.

Trusting all books is better than having no books

No, you didn't write it yourself, right?

.386c

cseg segment byte public use16
assume cs:cseg, ds:dseg, ss:sseg
start:
mov ax,dseg
mov ds,ax
lop:
xor si,si ; Clear si
mov word ptr ds:,0 ; Clear the value at memory ds:
cmp si, ; Is the value of si equal to the value at ds: (which is 0)?
je _exit ; If equal, exit the program
jmp lop ; Otherwise, loop and judge.
_exit:
mov ax, 4C00h
int 21h
cseg ends
;---------------------------------------------------
dseg segment byte public use16
db 0 dup(1024)
dseg ends
;----------------------------------------------
sseg segment stack
db 100h dup(?)
sseg ends
;----------------------------------------------
end start

The result of the program execution is an infinite loop, but according to the understanding that is the content of the memory address, it should exit the program normally. When debugging with tdb, it was found that the cmp instruction was compiled as cmp si,00C8, which means that was regarded by the compiler as the immediate number 0C8h, not the value stored at .

Another example:
mov ,ax ; XXXX is a numerical address, such as 1234h.
Compilation error: error A2001: immediate operand not allowed(Immediate operand is not allowed)
That is to say, the compiler regarded as an immediate number, not a memory address.
I changed it to mov ds:,ax and it compiled successfully. And it was correctly stored in that address.

Compiler version MASM6.11

No way, then it's just the same as the immediate number

Young man, find more problems in your own program!

Got it, in the source code, the form like is recognized by the MASM compiler as an immediate number (different from what the book says). Only and correspond to the memory value. However, in the debugger, is seen as a memory value, which is different from the source code.

The book is fine, just add a paragraph prefix and it's okay.

What the book says is correct. You can take a look at Wang Shuang's book. It's easy to understand.

I just tried it, and what's in the book is correct. The poster should find more reasons in himself

Reading books should capture their meaning, not just observe their appearance.

1. The BYTE declared in your data segment is used as WORD, so you need to add WORD PTR to restrict it.
2. mov memory, immediate data is acceptable, but it is not in the 8086 instruction set, so you need to add the.386 pseudo-instruction at the beginning of the program and declare the type of the memory variable, such as mov word ptr , 0.

The result of `mov ax,` on TASM is the same as that on MASM. The content of AX is 2000H.

Both reasons of the 12th floor are wrong, and the example given is correct.