# 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
```
• task1_1. Screenshot before the end of line17 and line19. • 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-0002H_， The segment address of stack is_ X-0001H_.

```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
```
• task1_2. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values • 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-0002H__， The segment address of stack is_ X-0001H___.

```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
```
• task1_3. Screenshot of the values of registers DS, CS and SS before the end of debugging to line17 and line19 • 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-0004H__， The segment address of stack is__ X-0002H__.

```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
```
• task1_4. After debugging to the end of line17 and before line19, observe the screenshot of register DS, CS and SS values • 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 Suppose that after the program is loaded, the segment address of the code segment is x, and the segment address of the data segment is X__ X+0002H__， The segment address of stack is_ X+0004H___.

Based on the practice and observation of the above four experimental tasks, summarize and answer:

• ① For the segment defined below, after the program is loaded, the actual memory space allocated to the segment is ceil(N/16)*16.
```      xxx segment
db N dup(0)
xxx ends
```
• 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_4 can be executed correctly. Because after changing to end, the program executes from the beginning, except task1_4. Other programs are not executed from scratch.

• source code
```assume cs:code

code segment
start:
mov ax,0b800h
mov ds,ax
mov bx,0f00h
mov cx,80

s:  mov [bx],0403h
loop s

mov ah, 4ch
int 21h
code ends
end start
```
• Screenshot of operation results • 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 dx,0
mov ax,data1
mov ds,ax

mov ax,data2
mov ds,ax

mov ax,data3
mov ds,ax
mov [bx],dl

inc bx

loop s

mov ax,4c00h
int 21h

code ends
end start
```
• Load, disassemble and debug screenshots in debug • Before data items are added in turn, check the original values of memory space data corresponding to logical segments data1, data2 and data3
debug command and screenshot • After adding in sequence, view the debug command and the original value of the memory space data corresponding to the logical segments data1, data2 and data3
screenshot • 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,10h

mov bx,0
mov cx,8

s:  push ds:[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. • source code
```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 it once before the program returns (i.e. after the execution of ine25 and line27)
Screenshot before execution) • What is the function of line19 in the source code? What is the purpose of the byte data in the data segment line4 in the source code?
line19 lowercase to uppercase; line4 set the font color.

• source code
```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 bx,0
mov cx,4

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. • source code
```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 si,0
mov di,0
mov bx,0
mov cx,5

mov sp,0

s: mov ax,[si]
mov es:[di],ax
mov ax,[si+2]
mov es:[di+2],ax

mov ax,[bx+20]
mov es:[di+5],ax
mov dx,0
mov es:[di+7],dx

push cx
mov cx,[20+10+bx]
mov es:[di+10],cx

div cx
pop cx

mov es:[di+0dh],ax

loop s

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 Posted by waradmin on Mon, 08 Nov 2021 07:08:58 -0800