指令系统_汇编程序复习.doc

Arm汇编语言语句分析：(1) 数据处理类指令MOV R0，R2 只能立即数之间SUB R0, R1, R2 只能立即数之间MOV R0, #0x101 立即数要满足一定要求MOV R0, R1, LSL #3 合法(2) Load/Store类指令LDR R0，R1, #4LDR R0, R1, R2, LSL #2!STR R0, R1, R2, LSL R3 移位的数只能是立即数(3) 杂类Load/Store指令LDRB R0, R1, #3LDRH R0, R1STRH R0, R1, R2, LSL #2 错 没有移位(4) 批量Load/Store指令Arm用的是Full Descending 堆栈STMFD R0!,R1,R2 = STMDB R0!,R1,R2LDMFD R0!,R1,R2 = LDMIA R0!,R1,R2(5) 状态寄存器访问指令MODE_SVC EQU0x13IRQ_DIS_BIT EQU0x80FIQ_DIS_BIT EQU0x40MOV R0, #0MOV R0, #MODE_ SVC:OR:IRQ_DIS_BIT:OR:FIQ_DIS_BITMSR CPSR_c, R0:OR: 为操作 把前后的数取或的关系ARM程序代码段分析： AREA Block, CODE, READONLY ; name this block of codenum EQU 20 ; Set number of words to be copied ENTRY ; mark the first instruction to callstart LDR r0, =src ; r0 = pointer to source block LDR r1, =dst ; r1 = pointer to destination block MOV r2, #num ; r2 = number of words to copy MOV sp, #0x400 ; set up stack pointer (r13)blockcopy MOVS r3,r2, LSR #3 ; number of eight word multiples BEQ copywords less than eight words to move ? STMFD sp!, r4-r11 ; save some working registersoctcopy LDMIA r0!, r4-r11 ; load 8 words from the source STMIA r1!, r4-r11 ; and put them at the destination SUBS r3, r3, #1 ; decrement the counter BNE octcopy ; . copy more LDMFD sp!, r4-r11 ; dont need these now - restore originalscopywords ANDS r2, r2, #7 ; number of odd words to copy BEQ stop ; No words left to copy ?wordcopy LDR r3, r0, #4 ; a word from the source STR r3, r1, #4 ; store a word to the destination SUBS r2, r2, #1 ; decrement the counter BNE wordcopy ; . copy morestop B END AREA BlockData, DATA, READWRITEsrc DCD 1,2,3,4,5,6,7,8,