Question

所以我怀疑为什么我们使用ADD AL，07H如果AL包含大于10的东西？对于ADD AL，07的解释是什么？这是代码。

MOV AH,01H  ;TAKE INPUT 
INT 21H

MOV BL,AL   ; SAVE VALUE OF AL, SO THAT IT CAN BE USED LATER
MOV CL,04H  
SHR AL,CL   ;SHIFT AL TOWARDS RIGHT BY 4 BITS

CMP AL,0AH  ;COMPARE IF AL HAS 10
JB DIGIT
ADD AL,07H
DIGIT: 
    ADD AL,30H  ;Add 30 to make HEX equivalent ASCII code
    MOV RES,AL

    AND BL,0FH
    CMP BL,0AH
    JB DIGIT1
    ADD BL,7H

DIGIT1: 
    ADD BL,30H
    MOV RES+1,BL

LEA DX,RES    ; display the result
    MOV AH,9
    INT 21H

谢谢。

Answer 1

您的程序（连续2次）将4位值转换为可显示字符。结果将遵循下表：

 0 -> "0"   "0" has ASCII 48 =  0 + 48
 1 -> "1"
 2 -> "2"
 3 -> "3"
 4 -> "4"
 5 -> "5"
 6 -> "6"
 7 -> "7"
 8 -> "8"
 9 -> "9"   "9" has ASCII 57 =  9 + 48
10 -> "A"   "A" has ASCII 65 = 10 + 48 + 7
11 -> "B"
12 -> "C"
13 -> "D"
14 -> "E"
15 -> "F"   "F" has ASCII 70 = 15 + 48 + 7

从这个表中你可以看到，当从9变为10时，ASCII不能很好地相互跟随。为了弥补这7个字符的间隙（它包含字符：;＆lt; =＆gt;？@ ）您使用了说明add al,7和add bl,7

使用汇编语言在屏幕上打印等效于字母数字字符的HEX ASCII

1 个答案: