请写出ADD -(R0), R1 指令、MOV @(R9)+,X(R0) 指令的指令格式和执行步骤。

ADD -(R0), R1 指令的指令格式为：

ADD  -(R0), R1

执行步骤如下：

1. 取出指令ADD -(R0), R1。
2. 将R0减1，得到新的R0值。
3. 读取新的R0指向内存中的值，记为M。
4. 将M与R1相加，将结果存入R1中。

MOV @(R9)+,X(R0) 指令的指令格式为：

MOV @(R9)+,X(R0)

执行步骤如下：

1. 取出指令MOV @(R9)+,X(R0)。
2. 读取R9指向的内存中的值，记为M。
3. 将M存入X(R0)指向的内存单元中。
4. 将R9加1，得到新的R9值。
5. 将R0加X，得到新的R0值。

相关问题

Use the following codes as the beginning of your program. Write a sort program to sort these ten numbers from largest to smallest, then save them to memory. For ARMSim#, the sorted number also need to be displayed in the console and LCD screen (use the Embest Board Plug‐In). MOV r0, #0x00002000 MOV r1, #18 STR r1, [r0],#4 MOV r1, #6 STR r1, [r0],#4 MOV r1, #2 STR r1, [r0],#4 MOV r1, #8 STR r1, [r0],#4 MOV r1, #16 STR r1, [r0],#4 MOV r1, #10 STR r1, [r0],#4 MOV r1, #14 STR r1, [r0],#4 MOV r1, #4 STR r1, [r0],#4 MOV r1, #20 STR r1, [r0],#4 MOV r1, #12 STR r1, [r0],#4，不要书写未知指令

Sure! Here's a possible solution in assembly language:

; Initialize registers
MOV r0, #0x00002000  ; Memory address to store sorted numbers
MOV r1, #18         ; First number
MOV r2, #6          ; Second number
MOV r3, #2          ; Third number
MOV r4, #8          ; Fourth number
MOV r5, #16         ; Fifth number
MOV r6, #10         ; Sixth number
MOV r7, #14         ; Seventh number
MOV r8, #4          ; Eighth number
MOV r9, #20         ; Ninth number
MOV r10, #12        ; Tenth number

; Sort numbers from largest to smallest using bubble sort algorithm
SORT_LOOP:
    MOV r11, #0           ; Flag to check if any swap was made in this iteration
    MOV r12, #9           ; Index of last element to compare
    CMP r12, #0           ; Check if only one element is left
    BEQ SORT_EXIT         ; If so, exit loop
SORT_INNER_LOOP:
    LDR r13, [r0, r12, LSL #2]  ; Load current element
    LDR r14, [r0, r12, LSL #2]  ; Load previous element
    CMP r13, r14          ; Compare current and previous elements
    BGE SORT_CONTINUE     ; If current element is larger, continue to next pair
    STR r13, [r0, r12, LSL #2]  ; If previous element is larger, swap them
    STR r14, [r0, r12, LSL #2]
    MOV r11, #1           ; Set swap flag
SORT_CONTINUE:
    SUBS r12, r12, #1     ; Move to previous pair
    BGT SORT_INNER_LOOP   ; Continue inner loop until all pairs are compared
    CMP r11, #1           ; Check if any swap was made in this iteration
    BNE SORT_EXIT         ; If not, exit loop
    MOV r11, #0           ; Reset swap flag
    MOV r12, #9           ; Reset index of last element to compare
    B SORT_LOOP           ; Continue outer loop
SORT_EXIT:

; Display sorted numbers in console and LCD screen
MOV r0, #0x00002000  ; Memory address of sorted numbers
MOV r1, #10         ; Number of elements to display
MOV r2, #0          ; Initial index
DISPLAY_LOOP:
    LDR r3, [r0, r2, LSL #2]   ; Load current element
    ; Display number in console (using SWI 0x06)
    MOV r0, #0x06      ; SWI number for console output
    MOV r1, r3         ; Value to display
    SWI 0               ; Call SWI
    ; Display number in LCD screen (using SWI 0x11)
    MOV r0, #0x11      ; SWI number for LCD screen output
    MOV r1, #0         ; X-coordinate of number
    MOV r2, r2, LSL #2 ; Y-coordinate of number (multiply by 4 for 32-bit memory access)
    ADD r2, r2, #0x80  ; Offset of LCD screen memory (0x08000000)
    MOV r3, r3         ; Value to display
    SWI 0               ; Call SWI
    ADDS r2, r2, #4    ; Move to next line
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    ADDS r2, r2, #4
    SUBS r1, r1, #1    ; Decrement element count
    BNE DISPLAY_LOOP   ; Continue loop until all elements are displayed

Note that this program uses the bubble sort algorithm to sort the numbers from largest to smallest. The sorted numbers are stored in memory starting at address 0x00002000, and then displayed in both the console and LCD screen using software interrupts (SWIs). The SWI numbers for console and LCD screen output are 0x06 and 0x11, respectively. The X-coordinate for the LCD screen output is always 0, and the Y-coordinate is incremented by 4 for each line.

Address EQU 0x40005000 ; 定义一个变量，地址为0x40005000 AREA Example,CODE,READONLY ; 声明代码段Example ENTRY ; 标识程序入口 CODE32 ; 声明32位ARM指令 START LDR R1,=Address ; R1 <- Address MOV R0,#10 ; R0 <- 10 储存数据的个数作为外层循环 STR R0,[R1] ; [R1] <- R0 MOV R2,#8 MOV R3,#10 MOV R4,#15 MOV R13,#0x0f0 STMFD SP!,{R2-R4} LDMFD SP!,{R5-R7} LDR R5,=MyData3 LDR R4,=MyData2 LDR R3,=MyData1 SUB R5,R5,#1 LOOP LDRB R6,[R5,#1]! LDRB R7,[R5,#1]! LDRH R8,[R4],#2 LDRH R9,[R4],#2 LDR R1,[R3],#4 LDR R2,[R3],#4 CMP R1,R2 ; R1与R2比较 STRHI R2,[R3] STRLS R1,[R3] BL LOOP MyData1 DCD 10,20,30,40,50,60,70,80,90 ;第1组数据 MyData2 DCW 10,20,30,40,50,60,70,80,90 ;第2组数据 MyData3 DCB 10,20,30,40,50,60,70,80,90 ;第3组数据 END

这段代码是 ARM 汇编语言，它定义了一个地址为 0x40005000 的变量 Address，并且使用了三组数据 MyData1、MyData2、MyData3。代码中包括了外层循环和内层循环，循环次数分别为 10 和 8。循环体内使用了一些数据处理指令，例如 LDR、STR、ADD、SUB、CMP 等。整个程序的作用是对三组数据进行一些处理，并且将结果存储回 MyData1 中。