汇编语言-字母字符转换

1． 题目：大小写字母字符互换

2． 要求：从键盘输入一个字符，如果该字符是回车符，直接退出程序，如果是小写字母，则转换为大写字母并显示；如果是大写字母，则转换为小写字母并显示；如果是非字母字符，则显示提示信息，并等待用户重新输入字符。

3． 提示： 首先判断是否是回车符，如果不是，则判断是否是大写字母或小写字母，如果是进行转换并输出，否则显示重新输入。大写字母ASCII码与小写字母ASCII之间相差20H，可以根据这个进行转换。

 1 ; Example assembly language program -- 
 2 ; Author:  karllen
 3 ; Date:    revised 5/2014
 4 
 5 .386
 6 .MODEL FLAT
 7 
 8 ExitProcess PROTO NEAR32 stdcall, dwExitCode:DWORD
 9 
10 INCLUDE io.h            ; header file for input/output
11 
12 cr      EQU     0dh     ; carriage return character
13 Lf      EQU     0ah     ; line feed
14 
15 .STACK  4096            ; reserve 4096-byte stack
16 
17 .DATA
18       promot      BYTE "Enter a char of letter ",cr,Lf,0
19       warning     BYTE "The char isn't a letter,enter again ",0
20       
21       answerLtoU  BYTE "The char is a lowercase,it's uppercase is "
22                   BYTE cr,Lf,0
23       answerUtoL  BYTE "The char is a uppercase,it's lowercase is "
24                   BYTE cr,Lf,0
25       char        BYTE 1 DUP(?)
26 .CODE
27 _start:
28       output   promot
29       input char,1
30       doGo:
31         mov bl,char
32         input char,1
33         cmp char,0dh
34         je  doWhCMP     ;cr deal the cr
35       jmp doGo
36       doWhCMP:
37               
38              cmp bl,41h
39              jl  inputAgain   ;the char < A,end and input again
40              cmp bl,5Ah
41              jle endUppertoL ;the char <= Z,so to judge it is or not lowercase
42                              ;the char is a uppercase
43        cmpLower:
44              cmp bl,61h     ;the char < a,end and input again
45              jl  inputAgain
46              cmp bl,7Ah     ;the char > z.end and input agian
47              jg  inputAgain
48              jmp endLowertoU ;the char is a lowercase
49                    
50        inputAgain:
51              output warning 
52              input  char,1
53              mov    bl,char
54              jmp    doGo
55        endUppertoL:
56              mov    al,bl
57              add    al,32             ；这里是十进制加减
58              mov    char,al
59              output answerUtoL
60              output char
61              jmp    endMain
62        endLowertoU:
63              mov    al,bl
64              sub    al,32
65              mov    char,al
66              output answerLtoU
67              output char
68        endMain:
69                        
70               INVOKE  ExitProcess, 0  ; exit with return code 0
71 
72 PUBLIC _start                   ; make entry point public
73 
74 END                             ; end of source code