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

Chương 8
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 8 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 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
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

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 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
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 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
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 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

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
CMP r3,#0
;
BGT
Loop
POP {LR}
BX LR
ENDP

END
Câu 10: tìm hiểu về stack
AREA sum,CODE
EXPORT __main
IMPORT countbitdiffer
ALIGN
ENTRY
__main PROC

area mydata,code,readwrite
align
ketqua dcd 0
end

mov r0,#2
BL quad
b endl
sq1

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

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

if i<=0


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
AREA myData,CODE,READWRITE
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

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

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 <- 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]
stop B stop
ENDP
AREA myData,CODE,READWRITE
align
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

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
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 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
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 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 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