C语言 将一维数组的索引转换为访问二维数组

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

对吗？double cij = C[i+j*n];

据我所知，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

下面的语句

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

是正确的。为了理解这一点，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

对于row = 1

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

从上面你不能有*(ptr*8 + row)它应该是*(ptr + row*8).

同样，如果ptr像double ptr[2][3]一样2D数组，则

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

同样，在您的情况下，有效的一个C[i + j*n]不C[i*n +j]

编辑：- 您有如下所示2D array

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 -->  

现在当你做a[1*n + 3])它的内部扩展方式

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 */

当你这样做时a[1+3*n])它的内部如何扩展

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 */

当您这样做时*(*(A+1) +3))它在内部扩展为

*(*(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 */

当你做A[1][3]这与上述情况相同。