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N-gram algorithm in x86 assembly assignment help

The assignment deals with implementing the n-gram algorithm using intel x86 assembly written for 32bit NASM. The program uses a function that takes two strings and returns the similarity between the given strings. The main function is written in C and calls the assembly function following the x86 calling convention. The main function loads the strings from a file and displays the string similarity result on the screen. Here is the solution to the assignment provided by our x86 assembly homework helpers.

N-Gram Algorithm for a 32-Bit NASM Program

N-gram algorithm in x86 assembly assignment help

#include #include int n_gram(char* str_1, int size_1, char* str_2, int size_2, int n); int main() { char buffer[512]; FILE *input; int n; int result; char string1[100]; char string2[100]; printf("Input filename?: "); fgets(buffer, 512, stdin); buffer[strlen(buffer) - 1] = 0; input = fopen(buffer, "rt"); if (input == NULL) { printf("Unable to open file!\n"); return 1; } while (fgets(buffer, 512, input) != NULL) { sscanf(buffer, "%d %s %s", &n, string1, string2); result = n_gram(string1, strlen(string1), string2, strlen(string2), n); printf("Similarity %d-gram between %s and %s: %d%%\n", n, string1, string2, result); } fclose(input); return 0; } SECTION .data SnS DD 0 ; length of intersection SuS DD 0 ; length of union SECTION .text global n_gram ; int n_gram(char* str_1, int size_1, char* str_2, int size_2, int n); n_gram: push ebp ; save frame pointer mov ebp, esp ; create new stack frame push ebx ; save registers on stack push esi push edi mov DWORD[SnS], 0 ; initialize length of intersection to zero mov DWORD[SuS], 0 ; initialize length of union to zero mov ecx, [ebp + 12] ; load size1 mov esi, [ebp + 8] ; load pointer to str1 loop1: cmp ecx, [ebp + 24] ; compare remaining chars with n jl endloop1 ; if remaining < n, end loop push ecx ; save ecx on stack inc DWORD[SuS] ; this is a new ngram, increment union length mov edi, [ebp + 16] ; load pointer to str2 mov edx, [ebp + 20] ; load size2 loop2: cmp edx, [ebp + 24] ; compare remaining chars with n jl endloop2 ; if remaining < n, end loop cmp esi, [ebp + 8] ; see if this is the first loop jne skip ; if not, skip inc DWORD[SuS] ; else, this is a new ngram, increment union length skip: mov ebx, 0 ; initialize index to 0 compare: mov al, [esi + ebx] ; load character from first string cmp al, [edi + ebx] ; compare with second string jne next ; if not equal, go to next ngram inc ebx ; advance to next char cmp ebx, [ebp + 24] ; compare current index with n jl compare ; repeat while i < n inc DWORD[SnS] ; we found an intersection, increment length dec DWORD[SuS] ; decrement union length since one value was repeated next: inc edi ; advance to next ngram in second string dec edx ; decrement remaining chars of second string jmp loop2 endloop2: inc esi ; advance to next ngram in first string pop ecx ; restore ecx dec ecx ; decrement remaining chars of first string jmp loop1 endloop1: mov eax, 100 ; load 100 to convert to percentages mul DWORD[SnS] ; multiply by intersection length mov edx, 0 ; clear edx for making division div DWORD[SuS] ; divide 100* SnS / SuS to get similarity pop edi ; restore registers pop esi pop ebx pop ebp ; restore frame pointer ret