I've 2D arrays as below,
float** src;
which is filled with some values.
float des[x][x];
x is equal to the number of rows in src array.
So basically i want to generate a static 2D array from a dynamic 2D array.
I tried to use memcpy(&des, src, x), But it gives wrong result.
Any suggestions??
Well, 2D array element's are stored in memory in a row, |x| == memory cell
1.row |x|x|x|x|x|x| 2.row |x|x|x|x|x|x| 3.row |x|x|x|x|x|x|
while every pointer of array of pointer can point on completely different address in memory. For example first pointer points on array on adress 100, second points on adress 248, third on array on adress 2.
3.row |x|x|x|x|x|x| |?|?|?| 1.row |x|x|x|x|x|x| |?|?|?|?| 2.row |x|x|x|x|x|x|
So you could use memcpy on every row of arrays separately, or copy them element by element.
Hope it helps.
There are 2 problems with your memcpy:
src rows are not necessarily continuousConsider:
#define x 2
float src_row_0[x] = { 0.0f, 0.1f };
float src_row_1[x] = { 1.0f, 1.1f };
float * src_rows[x] = { src_row_0, src_row_1 };
float ** src = src_rows;
There is no guarantee that rows are next to each other in memory. So you cannot copy bytes with single memcpy. Same applies even if you allocated srcrows with malloc.
You'll need to copy each row separately.
x is not enough to copy all bytesmemcpy copies number of bytes, not number of elements. Single float variable is usually more than 1 byte in size. Simply passing x to memcpy is not enough. You'll need to multiply each number of items by size of an single element.
Corrected copy would look something like this:
for(int i=0; i<x; ++i) {
memcpy(&dst[i], src[i], x * sizeof(float));
}
float**is not contiguous in memory. You can't use memcpy only once. Try to understand what**means. You can do this with memcpyxtimes