[英]Converting the index of 1D array to access 2D array
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 */
}
}
}
}
这段代码来自Computer_Organization_and_Design_5th
这样对吗? 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,这没有任何意义...
以下声明
double cij = C[i+j*n];
是正确的。 为了理解这一点,我们假定double ptr[3] = {1.5,2.5,3.5]
,其中ptr
是三个double变量组成的数组。 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
从上面不能有*(ptr*8 + row)
它应该是*(ptr + row*8)
。
同样,如果ptr
是2D
数组,例如double ptr[2][3]
则
ptr[row][col] == *( *(ptr + row) + col*8)
同样,在您的情况下有效的是C[i + j*n]
而不是C[i*n +j]
编辑 :-您有一个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,当您执行a[1*n + 3])
其内部如何扩展
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 */
而当你做a[1+3*n])
它在内部如何扩展
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 */
当您执行*(*(A+1) +3))
其内部扩展为
*(*(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 */
并且当您执行A[1][3]
,与上述情况相同。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.