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

Trang 1

Chương 8 Câu 1: Tìm số nguyên tố từ 100 đến 200

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

ALIGN ketqua DCD 0 end

soNguyento.s

AREA codene,CODE EXPORT soNguyenTo ALIGN

soNguyenTo PROC

;R0 so kiem tra PUSH {LR}

CMP r0,#0x02;xet n<2 BGE Continue

B KetThuc Continue MOVS r1,#0x02 ;r1 = i;

B CheckVongFOr

; if(n%i==0)break;

Loop SDIV r2,r0,r1 MLS r2,r1,r2,r0 CBNZ r2,LenhTiepTuc

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 2: Viet chuong trình truyen 8 đối số

Main.s

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

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

endp end

Cau3: dịch chuyển kí tự asci Main.s

AREA mycode,CODE,READONLY EXPORT main

import shiftEncryption ALIGN

ENTRY main proc LDR r0,=ketqua MOV r1,#1

BL shiftEncryption stop B stop

endp AREA myData,CODE,READWRITE ALIGN

ketqua DCB "ABCDEF$$HIHA",0 end

cau8_3.s

AREA codene, code export shiftEncryption align

shiftEncryption proc

;r0 string shift encryption

;r1 offset mov r2,#0; i = 0 -> n loop LDRB r3,[r0,r2];r3 = str[i]

CMP r3,#0 BEQ exit; neu '\0' exit CMP r3,#'A' ; kiem tra thuoc trong 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

cau 4: MOIVE PRICE main.s

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 0 end

cau8_4.s

AREA codene,code export moviePrice align

moviePrice proc

;r0 so tuoi dau vao

;r1 gia ve dau ra 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

import calculates ALIGN ENTRY main proc MOV r0,#3 mov r1,#5

BL calculates

Trang 2

LDR r1,=ketqua

STR r0,[r1]

stop B stop

endp

ALIGN

ketqua DCD 0

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 <= 0; exit

BLE exit

mov r2,r1;r2 = n-i

mov r3,#1; tinh tich P moi so hang

loopa

cmp r2,#0; r2 == 0 break;

BLE continue

mul r3,r3,r0

sub r2,r2,#1

B loopa

continue

add r4,r4,r3

sub r1,r1,#1

B loop

exit

mov r0,r4; chuyen S -> r0

pop {r3,r4,r5,lr}

bx lr

endp

end

cau 6: Viết chương trình tính tổng giai thừa từ o

-> 10

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 0 end

câu 7: đếm số bít 1

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 1 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 area mydata,code,readwrite align

ketqua dcd 0 end

cau 8 : đếm số bít khác nhau của 2 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

so2 DCD 0xF000000F result DCD 2 END

Cau8_8.s

AREA sum,CODE EXPORT countbitdiffer ALIGN

countbitdiffer PROC

;dàu vao r0,r1

;thuc hien xor r0,r1 de tinh ra bit khac nhau

; tra ve so bit khac nhau r0 PUSH {LR}

EOR r2,r0,r1

; so bit khac nhau tuong duong so bit 1 co trong r2

; dem so bit 1 trong 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

POP {LR}

BX LR ENDP END Câu 10: tìm hiểu về stack AREA sum,CODE EXPORT main IMPORT countbitdiffer ALIGN

ENTRY main PROC mov r0,#2

BL quad

b endl sq1

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 2 END

Trang 3

Cau 11: tính f = ……

Main.s

AREA codene,CODE,READONLY

EXPORT main

IMPORT cal

ALIGN

ENTRY

main PROC

mov r0,#1

mov r1,#1

BL cal

LDR r1,=result

STR r0,[r1]

stop B stop

ENDP

export const_a

export const_b

export const_c

align

const_a dcd 1

const_b dcd 2

const_c dcd 1

result DCD 223

END

Cau8_11.s

area codede,code

export cal

import const_a

import const_b

import const_c

align

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

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

LDR r1,=result STR r0,[r1]

stop B stop ENDP AREA myData,CODE,READWRITE align

result DCD 223 END

Cau8_12.s area codene,code export fibonacci align

fibonacci proc

;r0 = so fibonacci thu n CMP r0,#0

BXEQ lr CMP r0,#1 BXEQ lr mov r1,#1 ; i : 1-> 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

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]

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 <- length sub r0,r0,#1 mov r1,#0; r1 <- 0 looppalin

LDRB r2,[r4,r0];

LDRB r3,[r4,r1];

cmp r2,r3 MOVNE r0,#0 BXNE lr cmp r1,r0 BGT successpalin

add r1,r1,#1 sub r0,r0,#1

b looppalin successpalin mov r0,#1

bx lr endp strlen proc

;r0 string

;< length : r0 mov r2,#0 loop

LDRB r1,[r0],#1 cmp r1,#0 BEQ exit add r2,r2,#1

B loop exit

mov r0,r2

bx lr endp end

câu 14: viết đệ qui main.s

AREA codene,CODE,READONLY EXPORT main

IMPORT recursive ALIGN

ENTRY main PROC mov r0,#2 mov r1,#3

BL recursive LDR r1,=result STR r0,[r1]

result DCD 111 END

Cau8_14.s

area codene,code export recursive align

recursive proc

;tinh n^k

; n -> r0

; k-> 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

Trang 4

Chương 10

Câu 1: xóa một kí tự trong 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 ra 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

EXPORT main

IMPORT strcat

ALIGN

ENTRY

main proc

LDR r0,=chuoi

LDR r1,=chuoi1

BL strcat

stop B stop

endp

ALIGN

diachi DCD 0

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 4 tim kí tự trong chuỗi

Main.s

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 0

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 của 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 ra khoi chuoi int i =0;

for(;i<10;i++){

students[i].c1 = 'a';

students[i].c2 = 'b';

students[i].c3 = 'c';

students[i].score = i;

} total = totalScore();

return 0;

}

totalScore.s

area condene,code import students export totalScore align

totalScore proc push {lr}

LDR r0,=students; r0 = students[0]

mov r1,#0; r1 = totalScore mov r2,#0; r2 = i:0-> 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 của 4 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 ra 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 ra 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 r2,#10 isPal

CBZ r4,done SDIV r1,r4,r2 ;r1 = n/10 MLS r3,r1,r2,r4 ;r3 = n - r1*10

add r1,r1,r1,LSL #2 ;r1 = 5*sum add r1,r3,r1,LSL #1 ;sum = sum * 10 +r

mov r2,#10

BL isPal done MOV r1,r5 CMP r1,r0 BNE no

B exit

exit pop {lr,r4,r5}

bx lr endp end

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