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

Keywords: Assembly Language

## 4, Experimental conclusion

```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-2h and the segment address of the stack is X-1h.

```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-2h and the segment address of the stack is X-1h.

```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-4h and the segment address of the stack is X-2h.

```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 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+2h and the segment address of the stack is X+4h.

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 16Byte.

```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.    Only task1_4.asm can still operate normally.

reason:
End in addition to informing the compiler of the end of the program, it can also inform the compiler of where the entry of the program is. The pseudo instruction end describes the end of the program and the entry of the program. After compilation and connection, the program entry indicated by "end start" is converted into an entry address and stored in the description information of the executable file.
Without the pseudo instruction, the exe file is loaded into memory. First, point CS:IP to the first address of the program. And only Task1_ In the program of 4, the code segment is defined at the beginning, so CS:IP points to the first address of the instruction. task1_ 1-1_ The data segment is defined at the beginning of the program of 3, so CS:IP points to the first address of the data rather than the instruction, and the CPU will execute the machine instructions in the data segment.

• Assembly source code

```assume cs:code

code segment
start:
mov ax, 0b800h
mov ds, ax
mov bx, 0f00h
mov cx, 50h
mov ax, 0403h
s:  mov ds:[bx], ax
loop s

mov ah, 4ch
int 21h

code ends
end start
```
• Screenshot of operation results  The results are consistent.

• 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 ax, data1
mov ds, ax
mov bx, 0
mov cx, 0ah
s:  mov ax, ds:[bx]
mov ds:[bx+20h], ax
inc bx
loop s

mov ah,4ch
int 21h

code ends
end start
```
• Load, disassemble and debug screenshots in debug It is required to give the debug command and screenshot to view the original value of memory space data corresponding to logical segments data1, data2 and data3 before adding data items in turn And, after adding in turn, 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 es,ax
mov bx,0
mov si,14
s:mov ax,ds:[bx]
mov es:[si],ax
sub si,2
loop s

mov ah, 4ch
int 21h

code ends
end start
```
• Load, disassemble and debug screenshots in debug It is required to give a screenshot of the memory space corresponding to data segment data2 by using the d command before the program exits.  ```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?

Convert lowercase letters to uppercase letters

• xxx segment db N dup(0) xxx ends 1 2 3 what is the purpose of the byte data of data segment line4 in the source code?

8-bit color 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
mov si,0
s:mov al,ds:[0+si]
or al,20H
mov ds:[0+si],al
loop s
mov ah, 4ch
int 21h
code ends
end start
```
• Load, disassemble and debug screenshots in debug It is required to give a screenshot of the memory space corresponding to the data segment data by using the d command before the program exits. ```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
s1:mov ax,ds:[bx]
mov es:[si],ax
mov ax,ds:[bx]
mov es:[si],ax
loop s1
mov bx,20
mov cx,5
mov si,5
s2:mov ax,ds:[bx]
mov es:[si],ax
mov word ptr es:[si],0
loop s2

mov bx,30
mov cx,5
mov si,10
s3:mov ax,ds:[bx]
mov es:[si],ax
loop s3

mov bx,20
mov di,30
mov cx,5
mov si,13
s4:mov ax,ds:[bx]
mov dl,ds:[di]
div dl
mov es:[si],al
loop s4

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

• 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 (optional)

Posted by Dragonfly on Sun, 07 Nov 2021 16:55:29 -0800