我正在为一个类的项目工作，它按照量规的要求工作，不过我想知道是否有更好的方法来实现一些东西。我在另一个项目中为一个不必要的'mov'停靠了几点。这是问题1。
“If-else（34分）：写一个程序，要求用户输入一个数字。如果该数字小于5，你将说明，你将加5，并将其存储在一个变量中;如果数字大于5，你将说明，然后从中减去5并将其存储在一个变量中;如果数字是5，你将向它添加3并说明并将其存储在一个变量中。“ p>
org 100h
mov dx, offset start ;move start of string address 'start' into dx
mov ah, 09h
int 21h ;print the string stored at DS:DX
mov ah, 01h ;function for getting keyboard input
int 21h
sub al, 30h ;subtract 30h to store our number as hexadecimal
mov bl, al ;copying data to BL as the following commands manipulate the data
;at AL.
cmp bl, 5 ;BL = AL
jz ifZero ;jump to ifZero if BL = 5
jl ifLess ;jump to isLess if BL < 5
jg ifGreater ;jump to ifGreater if BL > 5
ifZero: ;direct console output function
mov ah, 06h
mov dl, 0Ah
int 21h
mov dl, 0Dh
int 21h ;print newline and character return
add bl, 03h ;add 3 to BL, BL = 8
mov temp, bl
mov dx, offset eq ;move start of string address 'eq' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
ifLess: ;direct console output function
mov ah, 06h
mov dl, 0aH
int 21h
mov dl, 0Dh
int 21h ;print newline and character return
add bl, 05h ;add 5 to BL
mov temp, bl
mov dx, offset less ;move start of string address 'less' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
ifGreater:
mov ah, 06h
mov dl, 0ah
int 21h
mov dl, 0dh
int 21h ;print newline and character return
sub bl, 05h ;subtract 5 from BL
mov temp, bl
mov dx, offset great ;move starting address of string 'great' into dx
mov ah, 09h
int 21h ;print string
jmp exit ;unconditional jump to end program
exit:
ret
temp db ?
start db "Please enter a number: $"
less db "Less than 5... adding 5 $"
great db "Greater than 5... subtracting 5 $"
eq db "Equal to 5... adding 3 $"
问题3： 反控制循环。程序将要求用户输入一个字符然后显示 带有标签五次的角色 例如： 输入一个字符：A 您输入了：A
org 100h
mov cx, 05h ;counter controlled loop, start as 5
LabelLoop:
mov dx, offset prompt ;move string offset to dx
mov ah, 09h ;function for printing string from dx
int 21h
mov ah, 01h ;function to read character from keyboard
int 21h
mov bl, al ;preserving character read by copying to BL
;as register data for AL will be changing
;due to various functions
mov ah, 06h ;function for direct console output
mov dl, 0ah
int 21h
mov dl, 0dh ;these just make the text appear on a new
;line
int 21h
mov dx, offset output ;move the memory offset of output to dx
mov ah, 09h ;printing another string
int 21h
mov ah, 02h ;function to write a character to console
;gets the value from DL
mov dl, bl ;so we copy BL to DL and print it
int 21h
jmp newLine ;we unconditionally jump to the newLine
;label and print a new line for the program
;to run again
newLine:
mov ah, 06h
mov dl, 0ah
int 21h
mov dl, 0dh
int 21h
loop LabelLoop ;we jump to LabelLoop and CX = CX - 1
mov dx, offset goodbye
mov ah, 09h
int 21h
ret
prompt db 'Enter a character: $'
output db 'You entered: $'
goodbye db 'Good bye!$'
所以，我对这些问题的疑问是，还有更好的方法吗？键盘输入存储在AL中，但是每次我对AH执行mov功能时，寄存器值都会改变，无论是字符串打印还是字符打印。为了避免变量（因为它不是需求的一部分）或将其分配给内存（我们还没有学习），我将数据移动到另一个寄存器。这对于任何一个程序来说都是不必要的'mov'吗？
编辑：我意识到在
之后AL = DLmov ah, 06h
mov dl, 0ah
int 21h ;AL = DL after execution
答案 0 :(得分：3)
cmp bl, 5 ;BL = AL jz ifZero ;jump to ifZero if BL = 5 jl ifLess ;jump to isLess if BL < 5 jg ifGreater ;jump to ifGreater if BL > 5 ifZero:
您应该做的第一项改进是利用在代码中落实的可能性。不要使用jz ifZero，而是要落实，因为相等是jl和jg之后唯一的状态。 ifEqual 对于此状态也是更正确的名称。
cmp bl, 5 ;BL = AL
jl ifLess ;jump to isLess if BL < 5
jg ifGreater ;jump to ifGreater if BL > 5
ifEqual:
第二个优化是摆脱CR和LF的所有直接控制台输出。您应该将这些内容合并到要打印的消息中。这样做也将无需使用mov bl, al将AL复制到BL（您特别要求）：
less db 13,10,"Less than 5... adding 5 $"
great db 13,10,"Greater than 5... subtracting 5 $"
eq db 13,10,"Equal to 5... adding 3 $"
这是另一个失败的机会：
jmp exit ;unconditional jump to end program
exit:
您的第二个计划也可以从这些建议中受益。