LDR R3, [R0, R2, LSL #2]什么意思

这也是一个IT类问题。这条指令是ARM架构汇编语言中的一条指令，它的含义是从内存地址(R0 + R2*4)处读取一个32位的值，然后将这个值存储到寄存器R3中。其中，LSL #2表示R2左移2位，相当于将R2乘以4，因为在ARM架构中，一个32位的字（word）占据4个字节（byte）的空间。因此，LDR R3, [R0, R2, LSL #2]可以理解为从(R0 + R2*4)地址处读取一个32位的字，并将其存储到R3寄存器中。

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.

利用ARM汇编编写程序：将内存中从 0x30000000 开始的 100 个字数据，其结果存于 R3、R2中

下面是一个利用ARM汇编实现的程序，用于将内存中从0x30000000开始的100个字数据相加，结果分别存储在R3、R2寄存器中：

LDR R2, =0x30000000 ; 将0x30000000的地址存入R2寄存器
MOV R3, #0 ; 将R3寄存器清零
loop:
LDR R1, [R2], #4 ; 从内存中读取一个字到R1寄存器中，并将R2寄存器加上4
ADD R3, R3, R1 ; 将R1寄存器中的值加到R3寄存器中
SUBS R0, #100 ; 将计数器减一
BNE loop ; 如果计数器不为零，则继续循环
MOV R2, R3, LSR #16 ; 将R3寄存器的高16位存入R2寄存器中
MOV R3, R3, LSL #16, ASR #16 ; 将R3寄存器的低16位存入R3寄存器中

解释一下上述代码：

首先将0x30000000的地址存入R2寄存器，将R3寄存器清零。接下来进入循环，使用LDR指令从内存中读取一个字到R1寄存器中，并将R2寄存器加上4，以便读取下一个字。然后使用ADD指令将R1寄存器中的值加到R3寄存器中，以便计算累加和。循环执行100次，即可将所有数据相加。

最后，使用MOV指令和LSR、LSL、ASR指令将R3寄存器中的值分别存储在R3、R2寄存器中。其中，MOV R2, R3, LSR #16表示将R3寄存器的高16位存储在R2寄存器中；MOV R3, R3, LSL #16, ASR #16表示将R3寄存器的低16位存储在R3寄存器中。

注意：以上代码仅适用于从地址0x30000000开始的100个字数据相加，并且结果需要分别存储在R3、R2寄存器中的情况。如果数据的起始地址和数量不同，或者结果需要存储在其他寄存器中，则需要进行适当的修改。此外，以上代码未考虑数据溢出的情况。