高分悬赏!关于汇编语言的一些问题!

1) Add the contents of six memory bytes andd store the sum in memory:

a) Define variable VAR in the data segment to be 6 bytes long with random numbers in each byte.
b) Define a one-byte variable ANSWER that is initialized to zero.
c) Add VAR[0] through VAR[5] (6 bytes in total)
d) Put the sum in ANSWER.
e) Use a looping and any addressing mode that you find acceptable.

2)Change question one to keep track each time the carry bit is set.

a) Use a new variable called ANSWER1.
b) Each time carry is set add one to ANSWER1

Note: The word-size sum of all the bytes is: ANSWER1 for the high byte, ANSWER for the low byte.

3. This program is to count the number of lower and upper case characters and store the result in memory.
a) Store a string that includes upper case, lower case and other ASCII characters at location STRING1.
b) Write a program that reads 20 bytes of the character data that are stored at location STRING1 and counts the number of upper case characters. And the number of lower case charactors.
c) Store this total in location UPPER and location LOWER.

4. Change question three to keep reading any number of data bytes until the sentinal value 0FF16 is encountered. Make sure that the last byte stored at location STRING1 has the value 0FF16.

5. Write a program that gets a byte of data from location DATA1.
a) If DATA1 has bit 0 set, subtract 2016 from the value.
b) If bit 7 is set, clear bits 4 through 7.
c) If both bit 0 and bit 7 are set, then toggle bits 0 through 3.

Hint: You may find the test instruction useful for this problem.

6. Write a program which returns to DOS if x is pressed. Any other key is to be echoed to the screen.
This interrupt waits for a key press and returns the keycode:
mov ah,00h
int 16h ;interrupt 16h, function 0h
Exit condition:
• AH returns the keyboard scan code
• AL returns the ASCII character code

Key Scan code ASCII code
A 1Eh 41h (61h)
X 2Dh 58h (78h)
X 01h 1Bh
应楼下要求,悬赏我提高了50分.
但是我要的是完整详细的答案,不是要翻译....所以不能给你分~但是还是谢谢你.
继续等待.

第5题有点迷糊...

感谢锟鹏展翅为我找出错误，谢谢！
现在可能没问题了呵呵...
有什么问题的话还请指出

1

assume cs:code, ds:datas

datas segment
var db 100, 77, 69, 201, 32, 148
answer db 0
datas ends

code segment
start:
mov bx, datas
mov ds, bx

mov cx, 6
mov bx, offset var
xor ax, ax
mov si, 0

next:
add al, [bx+si]
inc si
loop next

mov [answer], al

mov ax, 4c00h
int 21h
code ends
end start

2

assume cs:code, ds:datas

datas segment
var db 100, 77, 69, 201, 32, 148
answer db 0
answer1 db 0
datas ends

code segment
start:
mov bx, datas
mov ds, bx

mov cx, 6
mov bx, offset var
xor ax, ax
mov si, 0

next:
add al, [bx+si]
adc ah, 0
inc si
loop next

;mov word ptr [answer], ax
mov answer, al
mov answer1, ah

mov ax, 4c00h
int 21h
code ends
end start

3

assume cs:code, ds:datas

datas segment
string1 db 'Should Auld Acquaint|ance Be Forgot, And Never Brought To Mind?'
upper db 0
lower db 0
datas ends

code segment
start:
mov bx, datas
mov ds, bx
mov bx, offset string1
mov si, 0

nxt0:
cmp si, 20
jz done
mov al, byte ptr [bx+si]
cmp al, 65
jl nxt2
cmp al, 122
jg nxt2
cmp al, 90
jg nxt1
add [upper], 1
jmp nxt2
nxt1:
add [lower], 1
nxt2:
inc si
jmp nxt0
done:

mov ax, 4c00h
int 21h
code ends
end start

4

assume cs:code, ds:datas

datas segment
string1 db 'Should Auld Acquaintance Be Forgot, And Never Brought To Mind?',16h,0ffh
upper db 0
lower db 0
datas ends

code segment
start:
mov bx, datas
mov ds, bx
mov bx, offset string1
mov si, 0

nxt0:
mov cx, word ptr [bx+si]
cmp cx, 0ff16h
jz done
mov al, byte ptr [bx+si]
cmp al, 65
jl nxt2
cmp al, 122
jg nxt2
cmp al, 90
jg nxt1
add [upper], 1
jmp nxt2
nxt1:
add [lower], 1
nxt2:
inc si
jmp nxt0
done:

mov ax, 4c00h
int 21h
code ends
end start

5

这题题目可能有错吧，为了第2小问能减去2016，我把ax拿来操作了，其他几问的操作都是在ah里完成的：
assume cs:code, ds:datas

datas segment
data1 dw 1234h
datas ends

code segment
start:
mov bx, datas
mov ds, bx

mov ax, [data1]

test ah, 1
jz nxt0
sub ax, 2016
nxt0:
test ah, 80h
jz nxt1
and ah, 0fh
nxt1:
test ah, 81h
jz nxt2
mov bh, ah
not bh
and bh, 0fh
and ah, 0f0h
or ah, bh
nxt2:
mov data1, ax

mov ax, 4c00h
int 21h
code ends
end start

6

assume cs:code

code segment
start:

lp0:
mov ah, 0
int 16h
cmp al, 78h
jz done
cmp al, 58h
jz done
mov dl, al
mov ah, 2h
int 21h
jmp lp0
done:

mov ax, 4c00h
int 21h
code ends
end start

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