2012-03-08 43 views
1

我試圖創建的真值表爲函數F =(A & B)XOR(C & d')一個真值表。我正在使用英特爾彙編語言,整個代碼必須相當短。創建使用英特爾組件

,我依然會碰到的主要問題是與迴路操縱變量A,B,C和d創建真值表的所有可能的輸入。通常這樣做的方式就像變量都是4位二進制數的一位一樣,並且從0到15計數,實現每個可能的輸入。我只有4個寄存器堆棧和RAM存儲器來完成這個任務,因此我將每個變量分配給一個寄存器,並且我使用RAM來存儲計數器和表格的輸出。我想不出用循環做這件事的好方法,但我必須使用一個,因爲我使用的模擬器不會在代碼生效的情況下用代碼運行。 (這是256字節)

這裏是代碼,稱爲程序60的子程序是我想壓縮的那一部分。我不必實際輸出一個真值表,只是告訴它產生了多少「Minterm's」(輸出爲1)。

 ;MAIN 
    MOV AL,0 ;INITIALIZING VARIABLES 
    MOV BL,0 
    MOV CL,0 
    MOV DL,0 
    MOV [20],AL ;INITIALIZE CURRENT NUMBER OF MINTERMS WITH 0 
    MOV [21],AL ;INITIALIZE COUNTER WITH 0 
LOOP: CALL 30 
    CALL 60 
    MOV [21],CL 
    CMP CL,15 
    JS LOOP 
    CALL E0 
    MOV AL,[20] ;MOVE NUMBER OF MINTERMS TO AL 
    MOV [FF],AL ;OUPUT NUMBER OF MINTERMS TO VDU(LOCATIONS [C0] TO [FF]) 

     ;PROCEDURE 30, IMPLEMENTS THE GIVEN BOOLEAN FUNCTION 
    ORG 30 ;WRITE CODE BEGINNING AT [30] 
    PUSH DL ;CURRENT VALUE OF D 
    PUSH CL ;CURRENT VALUE OF C 
    PUSH AL ;CURRENT VALUE OF A, IMPLEMENTING FUNCTION WON'T CHANGE D 
    AND AL,BL ;PERFORMS (AL AND BL), STORES VALUE IN AL 
    NOT DL ;INVERSE OF DL STORED IN DL 
    AND CL,DL ;PERFORMS (CL AND DL), STORES VALUE IN CL 
    XOR AL,CL ;PERFORMS (AL XOR CL), STORES VALUE IN AL 
    MOV BL,[20] ;MOVES CURRENT NUMBER OF MINTERMS TO BL 
    ADD AL,BL ;ADD TO CURRENT NUMBER OF MINTERMS IF MINTERM RESULTED FROM FUNCTION 
    MOV [20],AL ;STORE NEW CURRENT NUMBER OF MINTERMS 
    MOV CL,[21] ;MOVE COUNTER TO CL 
    INC CL ;INCREMENT CL, INCREMENTING COUNTER 
    MOV [21],CL ;MOVE NEW VALUE OF COUNTER 
    POP AL ;RESTORE PREVIOUS VALUE OF D 
    POP CL ;RESTORE PREVIOUS VALUE OF C 
    POP DL ;RESTORE PREVIOUS VALUE OF A 
    RET 

     ;PROCEDURE 60, MANIPULATES VARIABLES TO IMPLEMENT FULL TRUTH TABLE 
    ORG 60 ;WRITE CODE BEGINNING AT [60] 
    MOV CL,[21] ;MOVE COUNTER VALUE TO CL REGISTER 
    CMP CL,1 ;NEED TO MAKE THIS INTO A LOOP 
    JZ FIRST 
    CMP CL,2 
    JZ SECOND 
    CMP CL,3 
    JZ FIRST 
    CMP CL,4 
    JZ FOURTH 
    CMP CL,5 
    JZ FIRST 
    CMP CL,6 
    JZ SECOND 
    CMP CL,7 
    JZ FIRST 
    CMP CL,8 
    JZ EIGHTH 
    CMP CL,9 
    JZ FIRST 
    CMP CL,10 
    JZ SECOND 
    CMP CL,11 
    JZ FIRST 
    CMP CL,12 
    JZ FOURTH 
    CMP CL,13 
    JZ FIRST 
    CMP CL,14 
    JZ SECOND 
    CMP CL,15 
    JZ FIRST 
    JMP LAST 
FIRST: 
    MOV AL,1 
    JMP LAST 
SECOND: 
    MOV AL,0 
    MOV BL,1 
    JMP LAST 
FOURTH: 
    MOV AL,0 
    MOV BL,0 
    MOV CL,1 
    JMP LAST 
EIGHTH: 
    MOV AL,0 
    MOV BL,0 
    MOV CL,0 
    MOV DL,1 
    JMP LAST 
LAST: 
    RET 

     FINISH: 
    ORG E0    ;WRITE CODE BEGINNING AT E0(OUTPUTS TO VDU) 
    DB "The number of Minterms in F is" ;OUTPUT STATEMENT 
    RET 

     END 
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你可以試試pcemu。我有一個叉只是爲了基本的x86彙編http://github.com/dwelch67/pcemu_samples – 2012-03-08 02:46:46

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使用一個寄存器並從0到15計數,使用AND和jz或jnz來確定每個位並從那裏開始。有很多解決方案,它不需要太多的代碼或註冊... – 2012-03-08 02:49:21

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如果這是一項家庭作業,請添加家庭作業標籤,如果這是x86程序集,請添加x86標籤。 – 2012-03-08 02:49:40

回答

2

如何簡單:

mov cl, 0 ; CL contains DCBA bits 
L: 
mov bl, cl ; BL contains DCBAb 
and bl, 101b ; BL contains 0C0Ab 
mov al, cl ; AL contains DCBAb 
shr al, 1 ; AL contains 0DCBb 
and al, 101b ; AL contains 0D0Bb 
and bl, al ; BL contains 0(C&D)0(A&B)b 
mov al, bl ; AL contains 0(C&D)0(A&B)b 
and bl, 1 ; BL contains 000(A&B)b 
shr al, 2 ; AL contains 000(C&D)b 
xor al, bl ; AL contains 000((C&D) XOR (A&B))b 
inc cl ; increment DCBA as a 4-bit number 
cmp cl, 16 ; all 16 combinations? 
jne L ; not yet, repeat 
0
xor cx,cx    ; init counter register 
repeatLoop: 
mov al,cx    ; bits a,b,c and d in lowest 4 nibbles 
call processtruthtable 
inc cx 
cmp cx,16    ; 4bits so 16 possibilities 
jl repeatLoop 
jmp exitProgram 
processtruthtable: 
.... 
.... 
.... 
.... 
ret 
exitProgram: 
// whatever OS you use, use the exit INT call 
for dos it's INT21/4C 
for linux it's INT80/01 
part after/is to be stored in AH 

糾正我,如果我錯了

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使用計數器與Alexey已提出的答案相同。順便說一下,我認爲你的意思是「最低4位」或「最低1位半字節」,因爲16位字只有4個半字節。 – 2018-02-17 13:07:44