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I have a column vector A which is 10 elements long. I have a matrix B which is 10 by 10. The memory storage for B is column major. I would like to overwrite the first row in B with the column vector A.
Clearly, I can do:
for ( int i=0; i < 10; i++ )
{
B[0 + 10 * i] = A[i];
}
where I've left the zero in 0 + 10 * i to highlight that B uses column-major storage (zero is the row-index).
After some shenanigans in CUDA-land tonight, I had a thought that there might be a CPU function to perform a strided memcpy?? I guess at a low-level, performance would depend on the existence of a strided load/store instruction, which I don't recall there being in x86 assembly?
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memmove() is similar to memcpy() as it also copies data from a source to destination.
The memcpy() function in C++ copies specified bytes of data from the source to the destination. It is defined in the cstring header file.
memcpy is only faster if: BOTH buffers, src AND dst, are 4-byte aligned. if so, memcpy() can copy a 32bit word at a time (inside its own loop over the length) if just one buffer is NOT 32bit word aligned - it creates overhead to figure out and it will do at the end a single char copy loop.
memcpy() itself doesn't do any memory allocations. You delete what you new , and delete[] what you new[] . You do neither new nor new[] . Both source and destination arrays are allocated on the stack and will be automatically deallocated when then go out of scope.
Short answer: The code you have written is as fast as it's going to get.
Long answer: The memcpy function is written using some complicated intrinsics or assembly because it operates on memory operands that have arbitrary size and alignment. If you are overwriting a column of a matrix, then your operands will have natural alignment, and you won't need to resort to the same tricks to get decent speed.
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