C语言 在 C 中分配矩阵
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allocate matrix in C
提问by Idan
i want to allocate a matrix.
我想分配一个矩阵。
is this the only option:
这是唯一的选择吗：
int** mat = (int**)malloc(rows * sizeof(int*))
for (int index=0;index<row;++index)
{
mat[index] = (int*)malloc(col * sizeof(int));
}
回答by jason
Well, you didn't give us a complete implementation. I assume that you meant.
好吧，你没有给我们一个完整的实现。我猜你是这个意思。
int **mat = (int **)malloc(rows * sizeof(int*));
for(int i = 0; i < rows; i++) mat[i] = (int *)malloc(cols * sizeof(int));
Here's another option:
这是另一种选择：
int *mat = (int *)malloc(rows * cols * sizeof(int));
Then, you simulate the matrix using
然后，您使用以下方法模拟矩阵
int offset = i * cols + j;
// now mat[offset] corresponds to m(i, j)
for row-major ordering and
用于行优先排序和
int offset = i + rows * j;
// not mat[offset] corresponds to m(i, j)
for column-major ordering.
用于列优先排序。
One of these two options is actually the preferred way of handling a matrix in C. This is because now the matrix will be stored contiguously in memory and you benefit from locality of reference. Basically, the CPU cache will a lot happier with you.
这两个选项之一实际上是在 C 中处理矩阵的首选方式。这是因为现在矩阵将连续存储在内存中，您可以从引用的局部性中受益。基本上，CPU 缓存会对您更满意。
回答by jamesdlin
The other answers already covered these, but for completeness, the comp.lang.c FAQ has a relevant entry:
其他答案已经涵盖了这些，但为了完整起见，comp.lang.c FAQ 有一个相关条目：
回答by abhay jain
what you can do is
你能做的是
int (*mat)[col];
mat=(int (*)[col])malloc(sizeof(*mat)*row);
and then use this new matrix as mat[i][j]
然后使用这个新矩阵作为 mat[i][j]
回答by Ana Betts
How about just:
怎么样只是：
int* mat = malloc(rows * columns * sizeof(int));
回答by Stéphan Kochen
You may also use calloc, which will additionally zero initialize the matrix for you. The signature is slightly different:
您也可以使用 calloc，它还会为您零初始化矩阵。签名略有不同：
int *mat = (int *)calloc(rows * cols, sizeof(int));
回答by Matthew Scharley
You cancollapse it to one call to malloc, but if you want to use a 2d array style, you still need the for loop.
您可以将其折叠为对 malloc 的一次调用，但是如果您想使用二维数组样式，您仍然需要 for 循环。
int** matrix = (int*)malloc(rows * cols * sizeof(int) + rows * sizeof(int*));
for (int i = 0; i < rows; i++) {
matrix[i] = matrix + rows * sizeof(int*) + rows * cols * sizeof(int) * i;
}
Untested, but you get the idea. Otherwise, I'd stick with what Jason suggests.
未经测试，但你明白了。否则，我会坚持杰森的建议。
回答by George
For a N-Dimensional array you can do this:
对于 N 维数组，您可以这样做：
int *matrix = malloc(D1 * D2 * .. * Dn * sizeof(int)); // Di = Size of dimension i
To access a array cell with the typical way you can do this:
要使用典型方式访问数组单元格，您可以执行以下操作：
int index = 0;
int curmul = 1;
int i;
int indexes = {I1, I2, ..., In}; // Ii = Index in dimension i
for(i = N-1; i >= 0; i--) {
index = index + indexes(i) * curmul;
curmul = curmul * Di;
}
(Note: didnt test now but should work. Translated from my Matlab code, but in Matlab index starts from 1, so i MAY made a mistake (but i dont think so))
（注意：现在没有测试但应该可以工作。从我的 Matlab 代码翻译，但在 Matlab 中索引从 1 开始，所以我可能犯了一个错误（但我不这么认为））
Have fun!
玩得开心！
