字符计数频率 - x86汇编

Question

我试图计算字符的出现在一个字符串。 我的代码是下面：

data segment 'DSEG' 
    text  db "This is a sentence.",0  ; string 
    textSize dw $ - text - 1    ; size of string, - 1 to account for null-termination character 
    freqTable dd 256 DUP(0) 
ends 'DSEG' 

code segment 'CSEG' 
start:       
mov ax, data  ; set segment registers 
mov ds, ax 
mov es, ax 
;--------------------------------------- 

sub cx, cx 
mov cx, textSize  ; number of times to loop 
L1: 
    mov ax, [OFFSET text + cx - 1] ; loop from back using cx, put character in ax 
    inc [OFFSET freqTable + 4*ax] ; increment table's index at the ascii value of character 
    LOOP L1 

;--------------------------------------- 
mov ax, 4c00h  ; return to OS 
int 21h 

ends 'CSEG' 
end start   ; set entry point 

我由DWORDS阵列，其中每个索引将代表一个字符。然后，我遍历字符串，并尝试以每个字符的ascii值增加数组。

不过，我得到一个wrong parameters错误，当我尝试在循环递增。我不确定是什么导致了这个错误。我猜我不能只是增加我想要的方式。我如何正确创建频率表？我错过了一些小事？

Answer 1

为了帮助您了解如何我创造下一个小程序与EMU8086（与组装兼容机）计算字符数：该程序要求用户输入一个文件名，打开文件，读取所有字符并计数，并关闭文件。

下一页图像显示它是如何工作：有频率的阵列（“freq_array”）256点的位置。每个位置都是相应字符的计数器，例如，位置65是'A'（chr（65））的计数器。

每当一个字符被从文件中读取，炭本身被用作偏移达到其计数器。例如，如果从文件读取char 48（'0'），则将数字48添加到数组偏移量（偏移+48），并且该位置增加。当文件结束时，所有的字符都被计算在内。

现在代码：

.model small 
.stack 100h 

;----------------------------------------- 

.data 

freq_array dw 256 dup(0) ;ARRAY OF FREQUENCIES OF EACH ASCII CHARACTER. 

msj   db 13,10,'Enter name of file: $' 

filename  db 99  ;MAX NUMBER OF CHARACTERS ALLOWED (98). 
      db ?   ;LENGTH (NUMBER OF CHARACTERS ENTERED BY USER). 
      db 99 dup(0) ;CHARACTERS ENTERED BY USER. END WITH CHR(13). 

filehandler dw ?   ;FILE HANDLER. 

the_char  db ?   ;CHAR READ FROM FILE. 

;----------------------------------------- 

.code 
start: 

;INITIALIZE DATA SEGMENT. 
    mov ax, @data 
    mov ds, ax     

    call get_source_file  ;GET FILE NAME. 
    call count_chars   ;FILL FREQ_ARRAY WITH FREQUENCIES OF CHARS. 

;WAIT FOR ANY KEY.  
    mov ah, 7 
    int 21h 

;FINISH PROGRAM. 
    mov ax, 4c00h 
    int 21h 

;----------------------------------------- 

get_source_file proc 
;DISPLAY MESSAGE. 
    mov dx, offset msj 
    mov ah, 9 
    int 21h  

;CAPTURE FILENAME FROM KEYBOARD.          
    mov ah, 0Ah 
    mov dx, offset filename 
    int 21h     

;CAPTURED STRING ENDS WITH CHR(13), BUT FILES REQUIRE 
;THE FILENAME TO END WITH CHR(0), SO LET'S CHANGE IT. 
    mov si, offset filename + 1 ;STRING LENGTH. 
    mov cl, [ si ]  ;MOVE LENGTH TO CL. 
    mov ch, 0    ;CLEAR CH TO USE CX. 
    inc cx    ;ONE MORE BYTE TO REACH CHR(13). 
    add si, cx   ;NOW SI POINTS TO CHR(13). 
    mov al, 0 
    mov [ si ], al  ;REPLACE CHR(13) BY 0. 

    ret 
get_source_file endp 

;----------------------------------------- 
;READ ALL CHARACTERS FROM FILE INCREASING THE COUNTER OF 
;EACH CHARACTER IN THE ARRAY OF FREQUENCIES. EACH CHARACTER 
;IS USED AS THE OFFSET OF ITS OWN COUNTER, EXAMPLE: THE 
;COUNTER FOR 'A' IS THE POSITION 65 OF FREQ_ARRAY. 

count_chars proc 
;OPEN FILE. 
    mov ah, 3dh   ;SERVICE TO OPEN FILE. 
    mov al, 0   ;OPEN AS READ ONLY. 
    mov dx, offset filename + 2 
    int 21h 
    mov filehandler, ax ;NECESSARY FOR OPERATIONS ON FILE. 

;COUNT CHARACTERS. 
reading: 
;READ ONE CHAR FROM FILE. 
    mov ah, 3fh   ;SERVICE TO READ FROM FILE. 
    mov bx, filehandler 
    mov cx, 1   ;HOW MANY BYTES TO READ. 
    mov dx, offset the_char ;WHERE TO STORE THE READ BYTES. 
    int 21h    

;CHECK END OF FILE. 
    cmp ax, 0 
    je end_reading  ;IF READ ZERO BYTES, FINISH. 

;INCREASE COUNTER. THE CHAR ITSELF IS BEEN USED AS INDEX: THE 
;COUNTER FOR CHAR 65 ('A') IS IN THE 65th POSITION OF THE ARRAY. 
    mov si, offset freq_array 
    mov al, the_char  ;USE CHAR AS OFFSET OF ITS OWN COUNTER. 
    mov ah, 0   ;CLEAR AH TO USE AX. 
    shl ax, 1   ;AX * 2, BECAUSE EVERY COUNTER IS 2 BYTES. 
    add si, ax   ;SI POINTS TO COUNTER POSITION. 
    inc [ word ptr si ] ;INCREMENT COUNTER FOR CURRENT CHAR. 
    jmp reading   ;REPEAT PROCESS. 

end_reading:   
;CLOSE FILE. 
    mov ah, 3eh   ;SERVICE TO CLOSE FILE. 
    mov bx, filehandler 
    int 21h 

    ret 
count_chars endp 

;----------------------------------------- 

end start 

希望这有助于你。

这是16位，这是因为阵列是DW。以使其与32位（阵列DD）兼容，改变下一行：

freq_array dd 256 dup(0) 

shl ax, 2    ;AX * 4, BECAUSE EVERY COUNTER IS 4 BYTES. 

inc [ dword ptr si ] ;INCREMENT COUNTER FOR CURRENT CHAR. 

Answer 2

X86不允许存储器通过使用AX的寻址和CX注册为你没有。但是，您可以使用EAX和ECX来实现此目的。
结束下一代码片段EAX [8,31]和ECX [16,31]将保持为零。

;--------------------------------------- 
xor eax, eax 
xor ecx, ecx 
mov cx, textSize  ; number of times to loop 
L1: 
mov al, [OFFSET text + ecx - 1] 
inc [OFFSET freqTable + eax*4] 
loop L1 
;---------------------------------------