Experiment 1,
1 assume ds:data, cs:code, ss:stack 2 3 data segment 4 db 16 dup(0) ; 16 byte units are reserved, and the initial values are 0 5 data ends 6 7 stack segment 8 db 16 dup(0) ;16 byte units are reserved, and the initial values are 0 9 stack ends 10 code segment 11 start: 12 mov ax, data 13 mov ds, ax 14 15 mov ax, stack 16 mov ss, ax 17 mov sp, 16 ; Set stack top 18 19 mov ah, 4ch 20 int 21h 21 code ends 22 end start① 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___. Note: ds=cs-2,ss=cs-1. Task 1-2 To program task1_2.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results.
1 assume ds:data, cs:code, ss:stack 2 3 data segment 4 db 4 dup(0) 5 data ends 6 7 stack segment 8 db 8 dup(0) 9 stack ends 10 code segment 11 start: 12 mov ax, data 13 mov ds, ax 14 15 mov ax, stack 16 mov ss, ax 17 mov sp, 8 18 19 mov ah, 4ch 20 int 21h 21 code ends 22 end start① 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__. Task 1-3 To program task1_3.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results
1 assume ds:data, cs:code, ss:stack 2 3 data segment 4 db 20 dup(0) 5 data ends 6 7 stack segment 8 db 20 dup(0) 9 stack ends 10 code segment 11 start: 12 mov ax, data 13 mov ds, ax 14 15 mov ax, stack 16 mov ss, ax 17 mov sp, 20 18 19 mov ah, 4ch 20 int 21h 21 code ends 22 end start① 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__. Tasks 1-4 To program task1_4.asm assembles and connects, loads and tracks debugging with debug, and answers questions based on the results. 1 assume ds:data, cs:code, ss:stack 2 code segment 3 start: 4 mov ax, data 5 mov ds, ax 6 7 mov ax, stack 8 mov ss, ax 9 mov sp, 20 10 11 mov ah, 4ch 12 int 21h 13 code ends 14 15 data segment 16 db 20 dup(0) 17 data ends 18 19 stack segment 20 db 20 dup(0) 21 stack ends 22 end start① 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__, Then, the segment address of the data segment is__ X+2__， The segment address of stack is_ X+4___. Tasks 1-5 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__ (N/16)*16__.
1 xxx segment 2 db N dup(0) 3 xxx ends 123② If the program Task1_ 1.asm, task1_ 2.asm, task1_ 3.asm, task1_ 4. In ASM, the pseudo instruction end start is changed to end, which program can still execute correctly? The reasons are analyzed and explained in combination with the conclusions obtained from practical observation. task1_4 can be executed, because after changing the end stat to end, the start does not match the end start, so the program cannot run from start, so the program runs sequentially from the beginning. Only program 4 can run as a program segment.
Experiment task 2 Write an assembly source program to realize 160 consecutive bytes to memory units b800:0f00 ~ b800:0f9f, and fill hexadecimal numbers repeatedly in turn According to 03 04. Tips: 1. In experiment task 3 of Experiment 1, experiments with the same effect were done. But at that time, the filling was realized through the f command of debug. This Second, programming is required. In debug, use the f command to batch fill in data to memory cells. -f b800:0f00 0f9f 03 04 Fill the memory unit interval b800:0f00 ~ b800:0f9f for 160 consecutive bytes, and fill hexadecimal data 03 and 04 repeatedly in turn. 2. When programming, pay attention to the problems of hexadecimal and byte order. After writing, run the program. If the result is different from that of experiment task 3 of Experiment 1 To, indicating that there is an error in programming.
1 assume cs:code 2 3 code segment 4 start: 5 mov ax,0b800h 6 mov ds,ax 7 mov bx,0f00h 8 mov cx,80 9 10 s: mov [bx],0403h 11 add bx,2 12 loop s 13 14 mov ah, 4ch 15 int 21h 16 code ends 17 end start
Operation results:
Experimental task 3 Code after completion:
1 assume cs:code 2 data1 segment 3 db 50, 48, 50, 50, 0, 48, 49, 0, 48, 49 ; ten numbers 4 data1 ends 5 6 data2 segment 7 db 0, 0, 0, 0, 47, 0, 0, 47, 0, 0 ; ten numbers 8 data2 ends 9 10 data3 segment 11 db 16 dup(0) 12 data3 ends 13 14 code segment 15 start: 16 mov bx,0 ; take bx Define as offset 17 mov dx,0 ; take dx Defined as register 18 mov cx,10 ; cx Defined as the number of summations 19 20 S: mov dx,0 ; Every time the register is cleared 21 mov ax, data1 22 mov ds ,ax 23 add dl, [bx] 24 25 mov ax, data2 26 mov ds ,ax 27 add dl, [bx] 28 29 mov ax, data3 30 mov ds ,ax 31 add [bx],dl 32 33 inc bx 34 loop S 35 36 mov ax,4c00h 37 int 21h 38 39 code ends 40 end start
requirement: ① The programming adds the data of logical segment data1 and logical segment data2 in turn, and the results are saved in logical segment data3. ② Load, disassemble and debug in debug. Before and after the data items are added in turn, view the three logical segments data1 respectively, Confirm that the memory space corresponding to data2 and data3 is added one by one to ensure that the result exists in the logical segment data3.
Screenshot of original data:
Execute the command
Results after operation:
Experimental task 4 It is known that the 8086 assembly source program task4.asm code fragment is as follows.
1 assume cs:code 2 3 data1 segment 4 dw 2, 0, 4, 9, 2, 0, 1, 9 5 data1 ends 6 7 data2 segment 8 dw 8 dup(0) 9 data2 ends 10 11 code segment 12 start: 13 mov ax, data1 14 mov ds, ax 15 mov ax, data2 16 mov ss, ax 17 mov sp, 16 18 19 mov bx, 0 20 mov cx, 8 21 s: 22 push [bx] 23 add bx, 2 24 loop s 25 26 mov ah, 4ch 27 int 21h 28 code ends 29 end startrequirement: ① Complete the program to store the eight word data in logical segment data1 in reverse order in logical segment b. ② After assembly and connection, load the program in debug and run it to line15. Before the program exits, use the d command to view the data corresponding to data segment data2 Memory space, confirm whether the subject requirements are met. Experimental results: it is found that the requirements of the topic are successfully completed
Experiment task 5 Use any text editor to enter the assembly source program task5.asm.
1 assume cs:code, ds:data 2 data segment 3 db 'Nuist' 4 db 2, 3, 4, 5, 6 5 data ends 6 7 code segment 8 start: 9 mov ax, data 10 mov ds, ax 11 12 mov ax, 0b800H 13 mov es, ax 14 15 mov cx, 5 16 mov si, 0 17 mov di, 0f00h 18 s: mov al, [si] 19 and al, 0dfh 20 mov es:[di], al 21 mov al, [5+si] 22 mov es:[di+1], al 23 inc si 24 add di, 2 25 loop s 26 27 mov ah, 4ch 28 int 21h 29 code ends 30 end startRead the source program, theoretically analyze the functions of the source code, especially line15-25, what are the functions realized by the loop, and understand the functions of each instruction line by line. Assemble and link the program to get the executable file, run and observe the results. Use the debug tool to debug the program and observe the results before the program returns, that is, after line25 and before line27.
What is the function of line19 in the source code?
Use bitwise and operations to capitalize letters
Modify the value of 5 byte units in line4, reassemble, link, run and observe the results.1 db 2,3,4,5,6 2 --> Change to: 3 db 5 dup(2) or db 5 dup(5)Based on observation, analyze and guess what the numerical function here is.
Here, the function of numerical value should be to change the color of letters.
Experimental task 6 It is known that the 8086 assembly source program task6.asm code fragment is as follows.
1 assume cs:code, ds:data, es:table 2 3 data segment 4 db '1975', '1976', '1977', '1978', '1979' 5 dw 16, 22, 382, 1356, 2390 6 dw 3, 7, 9, 13, 28 7 data ends 8 9 table segment 10 db 5 dup( 16 dup(' ') ) ; 11 table ends 12 13 code segment 14 start: 15 mov ax,data 16 mov ds,ax 17 mov ax,table 18 mov es,ax 19 20 mov bx,0 21 mov bp,0 22 mov si,20 23 mov cx,5 24 25 s: mov ax,ds:[bx] 26 mov es:[bp],ax 27 mov ax,ds:[bx+2] 28 mov es:[bp+2],ax 29 30 mov ax,ds:[si] 31 mov es:[bp+5],ax 32 mov word ptr es:[bp+7],0 33 34 mov ax,ds:[si+10] 35 mov es:[bp+10],ax 36 37 mov ax,ds:[si] 38 mov dl,ds:[si+10] 39 div dl 40 mov es:[bp+13],al 41 mov byte ptr es:[bp+14],0 42 43 add bx,4 44 add bp,16 45 add si,2 46 47 loop s 48 49 mov ah, 4ch 50 int 21h 51 code ends 52 end start
requirement: ① Complete the program, realize the title requirements, and write the year, income, number of employees and per capita income into the table section in a structured way. In the table, each row of data occupies 16 bytes in the logical segment table, and the byte size of each data is allocated as follows. Interim, between data Space spacing.
② After assembly and connection, load and debug the program in debug. Use u command, g command and d command flexibly and reasonably to display the beginning logic The data information of the segment table, and the data information of the segment table after the structured data is stored, confirm to realize the subject requirements. 
Results after running:
Tips: 1. This exercise task is not difficult, but cumbersome. You need to carefully sort out the layout of each data item in the table before you practice Move address). Then, the addressing mode and loop are flexibly applied to realize the movement of data between memory and memory. 2. When practicing, do not consider the decimal division, use the div command for integer division, and keep the quotient.