实验任务1

任务1-1

assume ds:data, cs:code, ss:stack


data segment
    db 16 dup(0)
data ends
stack segment
    db 16 dup(0)
stack ends
code segment
start:
    mov ax, data
    mov ds, ax

    mov ax, stack
    mov ss, ax
    mov sp, 16

    mov ah, 4ch
    int 21h
code ends
end start

1. 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = 076C，寄存器(SS) = 076D，寄存器(CS) = 076E
2. 假设程序加载后,code段的段地址是X,则,data段的段地址是X-2H, stack的段地址是 X-1H

Initial Program State

• ES and DS registers both point to the segment containing the PSP structure.
• CS equals value specified in the header, relocated by adding the start segment address to it.
• IP equals value specified in the header. Note, that unlike in COM executables, MZ programs don't start at offset 0x100.
• SS equals value specified in the header, relocated, just like CS.
• SP equals value specified in the header.
• AL is 0x00 if the first FCB in the PSP has a valid drive identifier, 0xFF otherwise.
• AH is the same as AL, but for the second FCB in the PSP.
• All other registers may, or may not be set to 0. You should consider them undefined.

Controlling the Segment Order^[1]

The assembler normally positions segments in the object file in the order in
which they appear in source code. The linker, in turn, processes object files in

the order in which they appear on the command line. Within each object file, the

linker outputs segments in the order they appear, subject to any group, class,
and .DOSSEG requirements.

任务1-2

assume ds:data, cs:code, ss:stack

data segment
    db 4 dup(0)
data ends

stack segment
    db 8 dup(0)
stack ends
code segment
start:
    mov ax, data
    mov ds, ax

    mov ax, stack
    mov ss, ax
    mov sp, 8

    mov ah, 4ch
    int 21h
code ends
end start

1. 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = 076C，寄存器(SS) = 076D，寄存器(CS) = 076E

2. 假设程序加载后,code段的段地址是X,则,data段的段地址是X-2H, stack的段地址是 X-1H

任务1-3

assume ds:data, cs:code, ss:stack

data segment
    db 20 dup(0)
data ends

stack segment
    db 20 dup(0)
stack ends
code segment
start:
    mov ax, data
    mov ds, ax

    mov ax, stack
    mov ss, ax
    mov sp, 20

    mov ah, 4ch
    int 21h
code ends
end start

1. 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = 076C，寄存器(SS) = 076E，寄存器(CS) = 0770

2. 假设程序加载后,code段的段地址是X,则,data段的段地址是X-4H, stack的段地址是 X-2H

任务1-4

1. 在debug中将执行到line17结束、line19之前,记录此时:寄存器(DS) = 076E，寄存器(SS) = 0770，寄存器(CS) = 076C

2. 假设程序加载后,code段的段地址是X,则,data段的段地址是X+2H, stack的段地址是 X+4H

任务1-5

1. 对于如下定义的段,程序加载后,实际分配给该段的内存空间大小是 $\lceil \frac{N}{16} \rceil$ Bytes。

   a physical segment can begin only at memory locations evenly divisible by 16^[2]

   由于 8086 特殊的寻址方式，段首地址只能是 16 的倍数。

2. 使用MASM，实际表现是四个程序均可以运行。使用TASM，无法通过编译。根据 Programmer's Guide^[3]:

   If you use full segment directives or prefer not to use .STARTUP, you must identify the starting instruction in two steps:

   1. Label the starting instruction.
   2. Provide the same label in the END directive.

   我认为不指定 starting instruction 的结果是未定义行为是合理的。

实验任务2

assume cs:code

code segment
start:
    mov ax, 0B8F0H
    mov ds, ax

    mov cx, 160
lb: 
    mov di, cx
    mov word ptr [di-2], 0403H
    dec cx
    loop lb

    mov ah, 4ch
    int 21h
code ends
end start

实验任务3

assume cs:code
data1 segment
    db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers
data1 ends

data2 segment
    db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0       ; ten numbers
data2 ends

data3 segment
    db 16 dup(0)
data3 ends

code segment
start:
    mov ax, data3
    mov es, ax

    mov cx, 10
    mov si, 0

lp: mov ax, data1
    mov ds, ax
    mov ax, ds:[si]
    mov es:[si], ax

    mov ax, data2
    mov ds, ax
    mov ax, ds:[si]
    add es:[si], ax

    inc si
    loop lp

    mov ah, 4ch
    int 21h
code ends
end start

before

after

实验任务4

assume cs:code

data1 segment
    dw 2, 0, 4, 9, 2, 0, 1, 9
data1 ends 

data2 segment
    dw 8 dup(?)
data2 ends

code segment
start:
    mov ax, data2
    mov es, ax

    mov ax, data1
    mov ds, ax

    mov si, 0
    mov di, 8*2
    mov cx, 8
lp: sub di, 2
    mov ax, word ptr ds:[si]
    mov word ptr es:[di], ax
    add si, 2
    loop lp
    
    mov ah, 4ch
    int 21h
code ends
end start

实验任务5

assume cs:code, ds:data
data segment
        db 'Nuist'
        db 2, 3, 4, 5, 6
data ends

code segment
start:
        mov ax, data
        mov ds, ax

        mov ax, 0b800H
        mov es, ax

        mov cx, 5
        mov si, 0
        mov di, 0f00h
s:      mov al, [si]
        and al, 0dfh
        mov es:[di], al
        mov al, [5+si]
        mov es:[di+1], al
        inc si
        add di, 2
        loop s

        mov ah, 4ch
        int 21h
code ends
end start

结果

退出前

line 19

如果字符为小写，则将这个字符转为大写。

line 4

颜色信息。参考VGA text mode。

实验任务6

assume cs:code, ds:data

data segment
    db 'Pink Floyd      '
    db 'JOAN Baez       '
    db 'NEIL Young      '
    db 'Joan Lennon     '
data ends

code segment
start:
    mov ax, data
    mov ds, ax

    mov di, 0
    mov cx, 5
lp: add byte ptr [di], 32
    add di, 10h
    loop lp

    mov ah, 4ch
    int 21h
code ends
end start

实验任务7

assume cs:code, ds:data, es:tb, ss:stk

data segment
    db '1975', '1976', '1977', '1978', '1979' 
    dw  16, 22, 382, 1356, 2390
    dw  3, 7, 9, 13, 28 
data ends

tb segment
    db 5 dup( 16 dup(' ') )  ;
tb ends

stk segment
    db 50 dup(0)
stk ends

code segment
start:
    mov ax, data
    mov ds, ax
    mov ax, tb
    mov es, ax

    ; field 1
    mov cx, 5 
_1: push cx
    mov cx, 4 
_2: mov al, [si]
    mov es:[di], al
    inc si
    inc di
    loop _2
    pop cx
    add di, 10h-4
    loop _1

    ; field 2
    mov di, 5
    mov cx, 5
_3: mov ax, word ptr [si]
    mov word ptr es:[di], ax
    add si, 2
    add di, 10h
    loop _3

    ; field 3
    mov di, 10
    mov cx, 5
_4: mov ax, word ptr [si]
    mov es:[di], ax
    ; signed div
    mov ax, [si-10]
    cwd
    idiv word ptr es:[di]
    mov es:[di+3], ax

    add si, 2
    add di, 10h
    loop _4
ext:
    mov ah, 4ch
    int 21h
code ends
end start

before

after