如何將具有多個變量的 c 程序轉換為匯編程序

Question

我必須將這個C程序轉換為匯編，但主寄存器不夠用，如何在匯編中添加更多變量？ （不從 C 導入它們）或者是否有一種簡單的方法可以在匯編中重現此代碼？

該程序的目標是比較最小的字符串（ S2 ）和最大的（ S1 ），並在字符串1中每次重復時將單詞的第一個字符的position保存在數組positions[]中

因此，如果

s1 = hello hello hellk s2 = he

預期的結果是

positions [0] = 0
positions [1] = 6
positions [2] = 12
positionsLen = 3

int main()
{

    #define MAX_LEN 100

    //input
    char s1[] = "hello hello hellk"; //17 char
    unsigned int lenS1 = sizeof(s1) - 1;
    char s2[] = "he"; //2 char
    unsigned int lenS2 = sizeof(s2) - 1;
    //output
    unsigned int positions[MAX_LEN];
    unsigned int positionsLen;

    ////-------------------------------------------------------

        int i, j, found;
        int stringLen, searchLen;

        stringLen = lungS1; // length of string
        searchLen = lungS2; // length of word to be searched

        positionsLen = 0; //positions[] index
        for (i = 0; i <= stringLen - searchLen; i++) //repeats the times the string s2 enters the string s1
        {
            //flag the word like found
            found = 1;
            for (j = 0; j < searchLen; j++) //scroll through the characters of the string 2
            {
                if (s1[i + j] != s2[j]) //I compare the two strings character by character (the number of times as the size of the string 2)
                //if the characters are different, the word is not found
                {
                    found = 0;
                    break;
                }
            }

            if (found == 1)
            //if the word was found from position i to position i + searchlen
            {
                positions[positionsLen] = i;
                positionsLen++;
            }
        }
return 0;

復制這個匯編程序的解決方案是什么？ 這是我的嘗試，但由於寄存器太小而導致溢出問題。

__asm
        {
            //dichiaro le variabili

            XOR AH, AH //i + j
            XOR AL, AL //s1
            XOR BH, BH //found 
            XOR BL, BL //scambio
            XOR ECX, ECX //i e j ma uso solo CX
            XOR DH, DH //volte
            XOR DL, DL //s2
            MOV DH, lungS1
            SUB DH, lungS2 //volte sarà la differenza tra le due stringhe e quindi le volte che il loop esterno si ripeterà
            
            MOV positionsLen, 0 //assegno la lunghezza del vettore



            MOV CX, 0 //imposto la i a 0
            ext_loop: nop //for con i
            MOV BH, 1 //FLAG TROVATO A 1
            MOV BL, CX //BL = CX (i)
            MOV CX, 0 //imposto la j a 0
            /////////////////////////////////////////////////////
            int_loop : //for con j
            //devo rendere AH == i+J
            //quindi
            MOV AH, BL //AH = i
            ADD AH, CX //AH = AH + CX(j)
            MOV AL, s1[AH]
            MOV DL, s2[CX]
            CMP AL, DL // CMP S1[i+j] == s2[j]
            je jump1//se sono uguali salta alla fine dell'if
            MOV BH, 0 //FLAG TROVATO A 0
            jmp int_loop_end //BREAK
            jump1 :
            INC CX //incremento j
            CMP CX, lungS2 //se j<lungS2 il ciclo si ripete
            jl int_loop
            /////////////////////////////////////////////////////
            int_loop_end :
            //tramite BL che non ho modificato nel ciclo interno
            //faccio tornare cx la variabile i.
            //(la j non mi servirà più fino al prossimo ciclo interno)
            MOV CX, BL
            MOV BL, 0
            //inizio l'if per verificare che sia stata trovata
            //la parola che inizia alla posizione i
            test BH, BH //se uguale a 0 salta
            jz jump2
            MOV positions[positionsLen], CX
            INC positionsLen
            jump2 :
            INC CX //INCREMENTO LA i
            CMP CX, DH //COMPARO LA i A volte
            jl ext_loop //SE MINORE CONTINUA
        }

Answer 1

自己寫匯編代碼復制一個C的代碼是非常困難和耗時的過程。 但是，您可以使用任何編譯器（ gcc 、 clang或msvc ）輕松執行此操作，它將生成C/C++代碼的匯編代碼。

假設您的C文件名為main.c ：

使用此命令生成名為main.s的匯編代碼：

gcc main.c -S

您可以使用優化標志優化最終的匯編代碼：

gcc -O3 -s -Og -masm=intel -march=native main.c -S