忽略一個字符串空格字符ASM

Question

試圖找到字符的字符串中的數量和不顧一切‘’空格字符

我有一個C++部分是通過琴絃ASM，這裏是我的ASM 工作得很好，唯一的事情就是太空人物也被計算在內。

stringLength PROC PUBLIC 
    PUSH ebp    ; save caller base pointer 
    MOV  ebp, esp  ; set our base pointer 
    SUB  esp, (1 * 4) ; allocate uint32_t local vars 
    PUSH edi 
    PUSH esi 
    ; end prologue 


    MOV esi, [ebp+8]   ;gets the string 
    xor ebx, ebx 

COMPARE: 
    MOV al, [esi + ebx]  
    CMP al, 0    ;compare character of string with 0 
    JE FINALE    ;if = to 0 go to end 
    INC ebx     ;counter 
    CMP al, ' '    ;compare with sapce 
    JE SPACE    ;go get rid of the space and keep going 
    INC al     ;otherwise inc al to next character and repeat 
    JMP COMPARE    

SPACE: 
    DEC ebx     ;get rid of the extra space 
    INC al 
    JMP COMPARE    ;goes back to compare 


FINALE: 

    MOV eax,ebx   ; bring back the counter 
    ADD esp, (2 * 4) ; clear the stack 
    POP esi 
    POP edi 
    MOV esp, ebp  ; deallocate locals 
    POP ebp    ; restore caller base pointer 
RET 

stringLength ENDP ; end the procedure 
END stringLength 

Answer 1

你正在做很多沒用的東西，沒有做任何事情來計算忽略空格。

1. 你並不真的需要安裝一個新的堆棧幀，這樣一個簡單的程序，你可以在被破壞寄存器所做的一切，或至多保存在堆棧的幾個寄存器;

2. 那inc al是沒有意義的 - 你正在遞增字符值，只是爲了在下一次循環迭代中丟棄它。

3. push fmt然後你立即清理堆棧？它有什麼意義？

4. mov ebx, 0 - 沒有人這樣做，零寄存器的慣用方法是xor ebx,ebx（指令編碼更緊湊）;

5. cmp al, 0鑑於你只對平等感興趣，你可以做test al, al（更緊湊）;

6. 您閱讀[ebp + 12]但從未實際使用它 - 是否應該是一個未使用的參數？

至於算法本身，你只需要保留一個單獨的計數器來計算非空格字符;實際上，您可以爲此保留ebx，並直接增加esi以迭代字符。例如：

xor ebx, ebx 
COMPARE: 
    mov al, [esi] 
    cmp al, ' ' 
    jne nonspace 
    inc ebx 
nonspace: 
    test al, al 
    jz FINALE 
    inc esi 
    jmp COMPARE 
FINALE: 

現在，這可以簡化進一步利用的事實eax將是返回值，並且您可以自由揍ecx和edx，所以：

stringLength PROC PUBLIC 
    mov ecx,[esp+4]   ; get the string 
    xor eax,eax    ; zero the counter 
compare: 
    mov dl,[ecx]   ; read character 
    cmp dl,' ' 
    jne nospace 
    inc eax     ; increase counter if it's a space 
nospace: 
    test dl,dl 
    jz end     ; go to end if we reached the NUL 
    inc ecx     ; next character 
    jmp compare 
end: 
    ret      ; straight return, nothing else to do 
stringLength ENDP ; end the procedure 

---

編輯：關於更新的版本

COMPARE: 
    MOV al, [esi + ebx]  
    CMP al, 0 
    JE FINALE 
    INC ebx 
    CMP al, " "  ; I don't know what assembler you are using, 
        ; but typically character values are in single quotes 
    JE SPACE 
    INC al   ; this makes no sense! you are incrementing 
        ; the _character value_, not the position! 
        ; it's going to be overwritten at the next iteration 
    JMP COMPARE    

SPACE: 
    INC eax   ; you cannot use eax as a counter, you are already 
        ; using it (al) as temporary store for the current 
        ; character! 
    JMP COMPARE 

Answer 2

我想我們需要使用整個eax寄存器來比較值。以這種方式：

; inlet data: 
; eax - pointer to first byte of string 
; edx - count of bytes in string 
; ecx - result (number of non-space chars) 
       push esi 
       mov ecx, 0 
       mov esi, eax 
@@compare: cmp edx, 4 
       jl @@finalpass 
       mov eax, [esi] 
       xor eax, 20202020h ; 20h - space char 
       cmp al, 0 
       jz @@nextbyte0 
       inc ecx 
@@nextbyte0: cmp ah, 0 
       jz @@nextbyte1 
       inc ecx 
@@nextbyte1: shr eax, 16 
       cmp al, 0 
       jz @@nextbyte2 
       inc ecx 
@@nextbyte2: cmp ah, 0 
       jz @@nextbyte3 
       inc ecx 
@@nextbyte3: add esi, 4 
       sub edx, 4 
       jmp @@compare 
@@finalpass: and edx, edx 
       jz @@fine 
       mov al, [esi] 
       cmp al, 20h 
       jz @@nextbyte4 
       inc ecx 
@@nextbyte4: inc esi 
       dec edx 
       jmp @@finalpass 
@@fine:  pop esi 
; save the result data and restore stack