Question

void dgem(int n, double *A, double *B, double *C)
{

    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; j++){

            double cij = C[i+j*n]; /* cij = C[i][j] */

            for(int k = 0; k < n; k++){
                cij += A[i+k*n] * B[k+j*n]; /*cij = A[i][k]*b[k][j]*/
                C[i+j*n] = cij; /* C[i][j] = cij */
            }
        }
    }

}

此代码来自Computer_Organization_and_Design_5th

是不是？ double cij = C[i+j*n];

据我所知，它应该是C[i*n + j]

int main(void){

    double A[4][4] = {1,2,3,4,
                      5,6,7,8,
                      9,10,11,12,
                      13,14,15,16};
    double * a = &A[0][0];

    int n = 4;
    printf("%f %f %f %f", *(*(A+1)+3), A[1][3], a[1*n + 3], a[1 + 3*n]); /* 
when i == 1 and j == 3 */

    return 0;

}

输出：

8.000000 8.000000 8.000000 14.000000

当我尝试使用gcc时，它没有意义......

Answer 1

以下声明

double cij = C[i+j*n];

是对的。要理解这一点让我们假设double ptr[3] = {1.5,2.5,3.5]其中ptr是三个双变量的数组。现在，您将如何访问ptr[0]或ptr[1]等。

 ----------------------------------------
|    1.5   |   2.5   |   3.5   |   4.5   |
-----------------------------------------
0x100     0x108    0x116     0x124     0x132 <-- lets say starting address 
LSB                                                 of ptr is 0x100
|
ptr

代表row = 1

ptr[row] == *(ptr + row * sizeof(ptr[row]))
2.5      == *(0x100 + 1*8)
2.5      == *(0x108)
2.5      == 2.5

从上面你不能*(ptr*8 + row) *(ptr + row*8)。

同样，如果ptr为2D数组，则为double ptr[2][3]，那么

ptr[row][col] == *( *(ptr + row) + col*8)

同样在您的情况下，有效的一个是C[i + j*n]而不是C[i*n +j]

修改： - 你有一个2D array，如下所示

double A[4][4] = { {1,2,3,4} , {5,6,7,8} , {9,10,11,12} , {13,14,15,16} };

和

double *a = &A[0][0];

现在看起来像

    A[0]          |   A[1]        |   A[2]           |   A[3]             |
  -------------------------------------------------------------------------
  | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |  16 |
  -------------------------------------------------------------------------
0x100 0x108...................0x156.......................0x156 (assume that 0x100 is starting address of A)
  A
  a
 LSB -->

现在当你a[1*n + 3])内部如何扩展

a[1*n + 3]) == *(a + (1*n + 3)  /*note that a is double pointer, it increments by 8 bytes
a[1*n + 3]) == *(0x100 + (1*4 + 3) *8)
a[1*n + 3]) == *(0x100 + 56)
            == *(0x156)
            ==  8 /* it prints 8.000000 */

当你做a[1+3*n])内部如何扩展

a[1+3*n]) == *(a + (1+3*n)*8)
a[1+3*n]) == *(0x100 + (1+3*4)*8)
a[1+3*n]) == *(0x100 + 96)
          == *(0x196)
          == 14 /* it prints 14.000000 */

当你执行*(*(A+1) +3))时，它会在内部扩展为

*(*(A+1) +3)) == *( *(0x100 +1*32) + 3*8) /* A+1 means increment by size of A[0] i.e 32 */ 
*(*(A+1) +3)) == *( *(0x132) + 24)
*(*(A+1) +3)) == *( 0x132 + 24 ) == *(156)
*(*(A+1) +3)) == 8 /*it prints 8.000000 */

当你执行A[1][3]时，与上述情况相同。

转换1D数组的索引以访问2D数组

1 个答案: