Ôn thi cuối kì vi điều khiển trường đại học bách khoa Đà nẵng

bài tập vi điều khiển được biên soạn theo chương trình chuẩn của Bộ giáo dục và đào tạo năm 2016. Bài tâp vi điều khiển giúp người học tiếp thu nhanh và cọ sát với đề thi cuối kỳ. Bài tập vi điều khiển trường đại học bách khoa đại học đà nẵng được biên soạn theo đề cương ôn thi cuối kỳ

Chương Main.s AREA mycode,CODE,READONLY EXPORT main IMPORT soNguyenTo ALIGN ENTRY main proc LDR r5,=ketqua mov r4,#100 loop CMP r4,#200 BGT stop MOV r0,r4 BL soNguyenTo cmp r0,#1 STREQ r4,[r5],#4 ADD r4,r4,#1 B loop stop B stop endp AREA myData,CODE,READWRITE Cau 2: Viet chuong trình truyen đối số Main.s AREA mycode,CODE,READONLY EXPORT main import extraargument ALIGN ENTRY main proc mov r0,#5 mov r1,#5 mov r2,#5 mov r3,#5 mov r4,#5 push {r4} mov r4,#6 push {r4} mov r4,#7 push {r4} mov r4,#8 push {r4} BL extraargument LDR r5,=ketqua STR r0,[r5] stop B stop endp AREA myData,CODE,READWRITE ALIGN ketqua DCD 10 end cau8_2.s AREA codene,CODE export extraargument align extraargument proc adds r0,r0,r1 adcs r0,r0,r2 adcs r0,r0,r3 pop {r4,r6,r5,r7} adds r0,r0,r4 adcs r0,r0,r5 adcs r0,r0,r6 adc r0,r0,r7 ;mov r1,#0 ;loop cmp r1,#4 ;BGE exit ;pop {r2} ;adcs r0,r0,r2 ;add r1,r1,#1 ;b loop exit bx lr Câu 1: Tìm số nguyên tố từ 100 đến 200 ALIGN ketqua DCD end soNguyento.s AREA codene,CODE EXPORT soNguyenTo ALIGN soNguyenTo PROC ;R0 so kiem tra PUSH {LR} CMP r0,#0x02;xet n n loop LDRB r3,[r0,r2];r3 = str[i] CMP r3,#0 BEQ exit; neu '\0' exit CMP r3,#'A' ; kiem tra thuoc khu vuc ki tu BLT continue CMP r3,#'Z' BLE shiftEN; CMP r3,#'a' BLT continue CMP r3,#'z' BGT continue shiftEN add r3,r3,r1 STRB r3,[r0,r2] continue add r2,r2,#1 B loop exit bx lr endp end B LenhBreak LenhTiepTuc ADDS r1,r1,#1 CheckVongFOr mov r3,r0;,LSR#1; ADD r3,r0,r0,LSR #31; n/2 CMP r1,r3,LSR #1 BLT Loop LenhBreak NOP ;if(i==n/2) ADD r3,r0,r0,LSR #31 CMP r1,r3,ASR #1 BNE KetThuc ; int result = 1; MOVS r0,#0x01 B Thoat KetThuc NOP MOVS r0,#0x00 Thoat NOP POP {LR} BX LR ENDP END cau 4: MOIVE PRICE main.s AREA mycode,CODE,READONLY EXPORT main import moviePrice ALIGN ENTRY main proc MOV r0,#13 BL moviePrice LDR r1,=ketqua STR r0,[r1] stop B stop endp AREA myData,CODE,READWRITE ALIGN ketqua DCD end cau8_4.s AREA codene,code export moviePrice align moviePrice proc ;r0 so tuoi dau vao ;r1 gia ve dau CMP r0,#12 MOVLE r0,#6 BLE exit CMP r0,#64 MOVLE r0,#8 BLE exit MOV r0,#7 exit bx lr endp end cau 5: tính tổng calculates Sn(a) = a + aa + aaa + … main.s AREA mycode,CODE,READONLY EXPORT main import calculates ALIGN ENTRY main proc MOV r0,#3 mov r1,#5 BL calculates LDR r1,=ketqua STR r0,[r1] stop B stop endp AREA myData,CODE,READWRITE ALIGN ketqua DCD end cau8_5.s area codene,code export calculates align calculates proc push {r3,r4,r5,lr} mov r3,#0; tong moi so hang mov r4,#0; tong S CMP r1,#0;keim tra truong hop n=0 MOVEQ r0,#0 loop cmp r1,#0; n r0 pop {r3,r4,r5,lr} bx lr endp end cau 6: Viết chương trình tính tổng giai thừa từ o > 10 AREA mycode,CODE,READONLY EXPORT main ALIGN ENTRY main proc BL sum LDR r1,=ketqua STR r0,[r1] stop B stop endp factorial proc CMP r0,#0 MOVEQ r0,#1 BXEQ lr mov r3,#1 loopfactorial cmp r0,#0 BLE exitfactorial mul r3,r3,r0 sub r0,r0,#1 B loopfactorial exitfactorial mov r0,r3; N! > r0 bx lr endp sum proc push {r4,r5,lr} mov r4,#10 mov r5,#0 loopsum cmp r4,#0 BLE exitsum mov r0,r4 BL factorial ADD r5,r0 sub r4,r4,#1 B loopsum exitsum mov r0,r5 pop {r4,r5,lr} bx lr endp AREA myData,CODE,READWRITE ALIGN ketqua DCD end câu 7: đếm số bít AREA mycode,code, readonly export main align entry main proc mov r0,#10 BL countbit1 LDR r1,=ketqua str r0,[r1] endp countbit1 proc ;r0 dem so bit mov r1,#32; i 0-> 32 mov r2,#0 ; result loopcount cmp r1,#0 BLE exitcount LSLS r0,r0, #1 adc r2,r2,#0 sub r1,r1,#1 B loopcount exitcount mov r0,r2 bx lr endp so2 DCD 0xF000000F result DCD END Cau8_8.s AREA sum,CODE EXPORT countbitdiffer ALIGN countbitdiffer PROC ;dàu vao r0,r1 ;thuc hien xor r0,r1 de tinh bit khac ; tra ve so bit khac r0 PUSH {LR} EOR r2,r0,r1 ; so bit khac tuong duong so bit co r2 ; dem so bit r2 MOV R0,#0 MOV R1,#0 MOV r3,#32; bien i =32; Loop LSLS r2,r2,#1 ; dich trai ADC r0,r0,#0 SUB r3,r3,#1 ; i = i-1 CMP r3,#0 ; BGT Loop POP {LR} BX LR ENDP END Câu 10: tìm hiểu stack AREA sum,CODE EXPORT main IMPORT countbitdiffer ALIGN ENTRY main PROC area mydata,code,readwrite align ketqua dcd end mov r0,#2 BL quad b endl sq1 cau : đếm số bít khác số main.s AREA sum,CODE EXPORT main IMPORT countbitdiffer ALIGN ENTRY main PROC LDR r0,=so1; LDR r0,[r0] LDR r1,=so2; LDR r1,[r1] BL countbitdiffer LDR r1,=result; STR r0,[r1] stop B stop ENDP AREA myData,DATA so1 DCD 13 push {lr} mul r0,r0 bl sq2 pop {pc} sq2 push {lr} mul r0,r0 pop {pc} quad push {lr} mul r0,r0 bl sq1 pop {lr} BX lr endl stop B stop ENDP AREA myData,DATA so1 DCD 13 so2 DCD 0xF000000F result DCD END if i n mov r2,#0 ; f(n-2) mov r3,#1 ; f(n-1) loop cmp r1,r0 BGT exit add r3,r3,r2 sub r2,r3,r2 add r1,r1,#1 b loop exit mov r0,r3 bx lr endp end cal proc ; r0 = x ; r1 = y ; r2 = ax2 ; r3 = bxy ; r4 = r2 + r3 + c push {r4,r5,lr} LDR r2,=const_a LDR r2,[r2] mul r2,r2,r0 mul r2,r2,r0 LDR r3,=const_b LDR r3,[r3] mul r3,r3,r0 mul r3,r3,r1 LDR r4,=const_c LDR r4,[r4] add r4,r4,r3 add r4,r4,r2 mov r0,r4 pop {r4,r5,lr} bx lr endp end cau 12: tính số Fibonacci main.s AREA codene,CODE,READONLY EXPORT main IMPORT fibonacci ALIGN ENTRY main PROC mov r0,#3 BL fibonacci cau 13: đối xứng chuỗi main.s AREA codene,CODE,READONLY EXPORT main IMPORT palindrome ALIGN ENTRY main PROC LDR r0,=chuoi BL palindrome LDR r1,=result STR r0,[r1] stop B stop ENDP AREA myData,CODE,READWRITE align chuoi DCB "abcdeedcba",0 result DCD 111 END Cau8_13.s area codene,code export palindrome align palindrome proc mov r4,r0 bl strlen; r0 r1 push {lr} cmp r1,#0 MOVEQ r0,#1 POPEQ {pc} push {r4} mov r4,r0 sub r1,r1,#1 bl recursive mul r0,r0,r4 pop {r4} pop {pc} endp end Chương 10 Câu 1: xóa kí tự chuối Main.c char str1[50] = "abc xyz abc chao den voi assembly"; char charRemove = 'a'; extern void removeChar(char *s,char c); int main(void){ //xoa mot ki tu khoi chuoi removeChar(str1,charRemove); return 0; } Removecharr.s AREA substring,CODE EXPORT removeChar ALIGN removeChar PROC ;r0 str[0] ;r1 char ;bao ton gia tri PUSH {r4,lr} mov r2,#0 loopend LDRB r2,[r0],#1; r2 = str1[i]; i++; CMP r2,#'\0' ; BEQ endloop CMP r2,r1 BNE loopend mov r3,r0 ;j=i swap LDRB r4,[r3],#-1;r4,str[j]; j ; STRB r4,[r3],#2 ;str1[j] = r4; j+=2; CMP r4,#'\0' bne swap B loopend endloop POP {r4,lr} BX LR ENDP Cau 2: nối chuối Main.s AREA mycode,CODE,READONLY EXPORT main IMPORT strcat ALIGN ENTRY main proc LDR r0,=chuoi LDR r1,=chuoi1 BL strcat stop B stop endp AREA myData,CODE,READWRITE ALIGN diachi DCD chuoi DCB "noi chuoi nay",0 chuoi1 DCB ", voi chuoi nay",0 end strcat.c void strcat(char *dst,char *src){ while(*dst) dst++; while(*dst++ = *src++); } Cau tim kí tự chuỗi Main.s AREA mycode,CODE,READONLY EXPORT main IMPORT search ALIGN ENTRY main proc LDR r0,=chuoi mov r1,#'a' BL search ldr r1,=ketqua str r0,[r1] stop B stop endp AREA myData,CODE,READWRITE ALIGN chuoi DCB "noi chuoi nay",0 ketqua dcd end searchChar.c char* search (char *s,char c){ char *p; for(;*s;s++){ if(*s == c) p=s; } return p; } Cau 5: tính tổng score students Main.c int total = 0; struct Student_T{ char c1; char c2; int score; char c3; } students[10]; extern int totalScore(); int main(void){ //xoa mot ki tu khoi chuoi int i =0; for(;i 10 add r0,r0,#4;r0 => students[i].score loop cmp r2,#10 BGE exit ;doc gia tri bo nho tai students[i].score LDR r3,[r0],#12 add r1,r1,r3 add r2,r2,#1 B loop exit mov r0,r1 pop {lr} bx lr endp end câu 6: tính max số main.c int ketqua = 0; int so1 = 11; int so2 = 22; int so3 = 33; int so4 = 44; extern int max4(int a,int b,int c,int d); int main(void){ //xoa mot ki tu khoi chuoi ketqua = max4(so1,so2,so3,so4); return 0; } Max4so.s area codene,code export max4 align max4 proc ;r0-r3 -> so1 - so4 ;return max -> r0 push {r4,lr} mov r4,r0 cmp r4,r1 MOVLT r4,r1 cmp r4,r2 MOVLT r4,r2 cmp r4,r3 MOVLT r4,r3 mov r0,r4 pop {r4,lr} bx lr endp end Cau 10_& số đối xứng Main.c int ketqua = 0; int so = 123321; extern int checks(int a); int main(void){ //xoa mot ki tu khoi chuoi ketqua = checks(so); return 0; } Cau10_7.s area codene,code export checks align checks proc ;r0 -> number ;return r0 = 1(true) | 0(false) push {lr,r4,r5} mov r4,r0 mov r5,#0 MOV r2,#10 ; sum = 0; isPal CBZ r4,done SDIV r1,r4,r2 ;r1 = n/10 MLS r3,r1,r2,r4 ;r3 = n - r1*10 mov r1,r5 ;r1 = sum add r1,r1,r1,LSL #2 ;r1 = 5*sum add r1,r3,r1,LSL #1 ;sum = sum * 10 +r mov r5,r1 mov r2,#10 SDIV r4,r4,r2 BL isPal done MOV r1,r5 CMP r1,r0 BNE no yes MOV r0,#1 B exit no mov r0,#0 exit pop {lr,r4,r5} bx lr endp end ;save sum ;r0 = n/10