通過更改起始代碼在c中進行基本乘法

Question

我需要採用以下代碼並使用mpz_替換使用我自己的代碼的調用。

void Product32(void *a, void *b, void *c, unsigned int wa, 
unsigned int ba, unsigned int wb, unsigned int bb, unsigned int 
*wc, unsigned int *bc){ 

mpz_t x,y,z; 
mpz_init(x); 
mpz_init(y); 
mpz_init(z); 

/* Cast a and b into short integers of size 32 bits */ 
unsigned int *int_a = (unsigned int *) a; 
unsigned int *int_b = (unsigned int *) b; 

/* Now int_a can be view as an array of words of size 32 
* bits */ 
/* Similarly for int_b */ 
//printf("%lu %lu \n", int_a[0], int_a[*sa - 1]); 
//printf("%lu %lu \n", int_b[0], int_b[*sb - 1]); 

mpz_import(x, wa, ORDER, WORDBYTES, ENDIAN, NAILS, a); 
mpz_import(y, wb, ORDER, WORDBYTES, ENDIAN, NAILS, b); 
mpz_mul(z,x,y); 

c = mpz_export(c, wc, ORDER, WORDBYTES, ENDIAN, NAILS, z); 
} 

我遇到的問題是，我不明白是什麼mpz_import或mpz_export完成的，而我尋找一個問題的答案已經來到了空。

我也覺得我的類型是完全錯誤的。

我忘了調用Product32的主函數，因爲我知道問題不存在;上面的代碼工作，下面的代碼不會。

這是我有：

/* Since we are working with string of potentially different lengths, 
first we need to be able to make the two strings of equal length. */ 

int makeEqualLength(int arr1[], int arr2[]) 
{ 
int len1 = sizeof(arr1); 
int len2 = sizeof(arr2); 
int i; 
if (len1 < len2) 
{ 
for (i = 0 ; i < len2 - len1 ; i++) 
    arr1[i] = arr1[i+1]; 
    arr1[0] = 0; 
return len2; 
} 
else if (len1 > len2) 
{ 
for (i = 0 ; i < len1 - len2 ; i++) 
    arr2[i] = arr2[i+1]; 
    arr2[0] = 0; 
} 
return len1; // If len1 >= len2 
} 


void Product32(void *a, void *b, void *c, unsigned int wa, 
unsigned int ba, unsigned int wb, unsigned int bb, unsigned int 
*wc, unsigned int *bc){ 

/* Cast a and b into short integers of size 32 bits */ 
unsigned int *int_a = (unsigned int *) a; 
unsigned int *int_b = (unsigned int *) b; 
unsigned int *int_c = (unsigned int *) c;  

/* Now int_a can be view as an array of words of size 32 
* bits */ 
/* Similarly for int_b */ 
//printf("%lu %lu \n", int_a[0], int_a[*sa - 1]); 
//printf("%lu %lu \n", int_b[0], int_b[*sb - 1]); 

int n = makeEqualLength(int_a, int_b); 
unsigned int i,j,k; 
double p; 

for (k = 0; k < n; i++){ 
int_c[k] = 0; 
} 
for (i = n - 1; i >= 0; i--){ 
double d = 0; 
for (j = n - 1; j >= 0; j--){ 
    p = (int_a[i]) * (int_b[j]) + int_c[i + j] + d; 
    int_c[i + j] = p % 32; 
    int_c = p/32 
    } 
int_c[i + n] = d; 
} 
} 

Answer 1

什麼你要找的是The GNU Multiple Precision Arithmetic Librarydocumentation。你會在那裏找到mpz_*函數的定義。

具體而言，mpz_import和mpz_export。他們完成的任務（將mpz_t變量轉換爲二進制數據的任意字）可以在這裏完全描述，它應該可以幫助你。