# Experiment 2 compilation and debugging of assembly source program of multiple logic segments

```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 In debug, execute until the end of line17 and before line19. Record this time: register (DS) =_ 076A___， Register (SS) =_ 076B___， Register (CS) =_ 076C___

② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X_ X-2___， The segment address of stack is_ X-1___.

```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 In debug, execute until the end of line17 and before line19. Record this time: register (DS) =_ 076A___， Register (SS) =_ 076B___， Register (CS) =_ 076C___

② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X_ X-2___， The segment address of stack is_ X-1___.

Because the memory space occupied by the program is continuous, and the segment size accounts for an integer multiple of 16 bytes, 4 and 8-byte segments will also occupy 16 bytes, so the result is the same as task1_1 consistent.

```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 In debug, execute until the end of line17 and before line19. Record this time: register (DS) =_ 076A___， Register (SS) =_ 076C___， Register (CS) =_ 076E___

② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X_ X-4___， The segment address of stack is_ X-2___.

Because the memory space occupied by the program is continuous, and the segment size accounts for an integer multiple of 16 bytes, the 20 byte segment will also occupy 32 bytes, so the result is this.

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

data segment
db 20 dup(0)
data ends

stack segment
db 20 dup(0)
stack ends
end start``` ① In debug, execute until the end of line9 and before line11. Record this time: register (DS) =_ 076C___， Register (SS) =_ 076E___， Register (CS) =_ 076A___

② Assuming that the segment address of the code segment is X after the program is loaded, the segment address of the data segment is X_ X+2___， The segment address of stack is_ X+4___.

Because the memory space occupied by the program is continuous, and the segment size accounts for an integer multiple of 16 bytes, the 20 byte segment will also occupy 32 bytes, and because the sequence of code segment data segment stack segment changes, the result is this.

```xxx segment
db N dup(0)
xxx ends```

① For the segment defined below, after the program is loaded, the actual memory space allocated to the segment is_ ceil(N/16)*16___.

② If the program Task1_ 1.asm, task1_ 2.asm, task1_ 3.asm, task1_ 4. In ASM, if the pseudo instruction end start is changed to end, which program can still be executed correctly. The reasons are analyzed and explained in combination with the conclusions obtained from practical observation.

task1_1.asm, task1_2.asm, task1_3.asm cannot be executed because they are all the beginning of the data segment. Replacing end start with end means that the program does not specify the start position and will start execution from the beginning, and the data in the execution data segment will naturally make an error. Only task1_4.asm can be executed because it starts with a code snippet.

Assembly source code

```assume cs:code
code segment
start:
mov ax,0b800h
mov ds,ax
mov bx,0f00h
mov cx,80
s:
mov [bx],0403h
inc bx
inc bx
loop s

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

Screenshot of operation results Complete assembly source code

```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 cx,10
mov bx,0
mov dx,0
s:
mov ax,data1
mov ds,ax
mov dl,[bx]
mov ax,data2
mov ds,ax
mov ax,data3
mov ds,ax
mov [bx],dl
inc bx
loop s
mov ah,4ch
int 21h
code ends
end start```

Load, disassemble and debug screenshots in debug  Before adding data items in turn, check the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3 After adding in sequence, view the debug command and screenshot of the original value of memory space data corresponding to logical segments data1, data2 and data3 Complete assembly source code

```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,data1
mov ds,ax
mov ax,data2
mov ss,ax
mov sp,16
mov cx,8
mov bx,0
s:
push [bx]
loop s
mov ah, 4ch
int 21h
code ends
end start```

Load, disassemble and debug screenshots in debug Before the program exits, use the d command to view a screenshot of the memory space corresponding to data segment data2. ```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
loop s

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

Screenshot of operation results Use the debug tool to debug the program, and use the g command to execute the screenshot before the program returns (i.e. after ine25 and before line27) What is the function of line19 in the source code?

A: convert lowercase letters to uppercase letters.

and al, 0dfh, where the conversion of 0dfh to binary is 1101 1111(B),

ASCII codes of capital letters are 0100 0001 ~ 0101 1010,

Lower case ASCII codes are 0110 0001 ~ 0111 1010,

After the ASCII code and odfh and of lowercase letters, the 1 of the third bit will become 0, which is converted from lowercase letters to uppercase letters.

What is the purpose of the byte data in the data segment line4 in the source code?

A: set the color of the display content.

Change the fourth line:  ```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 cx,4
mov bx,0
s:
mov al,[bx]
or al,20h
mov [bx],al
loop s
mov ah, 4ch
int 21h
code ends
end start```

Load, disassemble and debug screenshots in debug Before the program exits, use the d command to view a screenshot of the memory space corresponding to the data segment data. ```assume cs:code, ds:data, es:table

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

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

code segment
start:
mov ax,data
mov ds,ax
mov ax,table
mov es,ax
mov cx,5
mov bx,0
mov si,0
mov di,0
mov dx,0
s1:
mov ax,[bx]
mov es:[si],ax
mov ax,[bx+2]
mov es:[si+2],ax
loop s1
; mov bx,20
mov si,5
mov cx,5
s2:
mov ax,[bx]
mov es:[si],ax
mov ax,0
mov es:[si+2],ax
loop s2
;mov bx,30
mov si,10
mov cx,5
s3:
mov ax,[bx]
mov es:[si],ax
loop s3
mov si,0
mov cx,5
s4:
mov ax, es:[si+5]
mov dx, es:[si+7]
div word ptr es:[si+10]
mov es:[si+13],ax
loop s4
mov ah, 4ch
int 21h
code ends
end start```

View screenshot of original data information of table segment Run in debug until the program exits, use the d command to view the screenshot of the memory space corresponding to the table segment, and confirm whether the information is structurally written to the specified memory as required 5, Experimental summary

1.or 20h and and 0dfh commands are very practical. The former can convert uppercase letters to lowercase, and the latter can convert lowercase letters to uppercase. The combination of the two can be used flexibly to easily realize the mixed format output of uppercase and lowercase letters.

2. In order to prevent the register from being "polluted" by unknown data error when calling a register value, it is recommended to set the register that needs to be used but does not have a specific value to 0 in the assignment stage.

Posted on Sat, 06 Nov 2021 22:20:00 -0400 by WhiteHawksan