Fast Array Element Copy?

void* memcpy( void* dst, void* src, size_t sizeToCopy);

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Original post by DIB I am trying to copy elements within an array. The array is huge, so I need it to be fast. Here's what I am currently doing... for(int i = 0; i < totalArraySize; i+=5) { array[i+1] = array; array[i+2] = array; array[i+3] = array; array[i+4] = array; } Is there any memcpy-like function that can do this? I would love to know how to unroll this and make it faster. Any suggestions?? [edited for formatting]

int tmp;for(int i = 0; i < totalArraySize; i++){	tmp = array;	array[++i] = tmp;	array[++i] = tmp;	array[++i] = tmp;	array[++i] = tmp;}

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Original post by DIB [Now, sure, you can say I should modify the display function, etc, but this is a slight simplification of the problem, so trust me, if I want it pretty, I have to copy the elements.]

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I'm not a very trusting person. This is nothing a very minor rework couldn't solve, unless the code that relies on it is intentionally obfuscated.

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Original post by me22 If you're using C, use memcpy. If you're using C++, use std::copy. std::copy is usually implemented by your library writer to turn into a call to memcpy whenever possible, but it won't use memcpy in a case where you have a non-POD type, which would really screw you over.

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Original post by Zahlman Anyway, it looks to me as if you are already unrolling things - incorrectly, missing the elements with indices divisible by 5 :s

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Original post by DIB Ok, the reason I am doing this is: Each set of n array positions (5, in this case) starts out with a bunch of data in them. I have to move through the data, manipulating things, doing calculations, etc. When I am done, the result is in the first of the 5 bins. Now, I could just leave the result there, eliminating this problem. However, for presentation reasons, I need to copy it to the rest of the bins so the next function, which plots the data, looks prettier.

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If you actually want to copy a buffer to some overlapping location you should use memmove instead. memcpy is only guaranteed for non-overlapping buffers. MSDN says std::copy is defined to shift elements to the left. I'm not sure what that means offhand.

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Because the algorithm copies the source elements in order beginning with the first element, the destination range can overlap with the source range provided the _First position of the source range is not contained in the destination range. copy can be used to shift elements to the left but not the right, unless there is no overlap between the source and destination ranges. To shift to the right any number of positions, use the copy_backward algorithm.

std::vector< int > example( ... );<strike>std::copy          ( example.begin() + 1 , example.end()     , example.begin()     ); //shifts everything left 1 element  - <font color="green">OK</font>std::copy          ( example.begin()     , example.end() - 1 , example.begin() + 1 ); //shifts everything right 1 element - <font color="red">NOT OK</font>std::backwards_copy( example.begin()     , example.end() - 1 , example.begin() + 1 ); //shifts everything right 1 element - <font color="green">OK</font>std::backwards_copy( example.begin() + 1 , example.end()     , example.begin()     ); //shifts everything left 1 element  - <font color="red">NOT OK</font></strike>std::copy         ( example.begin() + 1 , example.end()     , example.begin()     ); //shifts everything left 1 element  - <font color="green">OK</font>std::copy         ( example.begin()     , example.end() - 1 , example.begin() + 1 ); //shifts everything right 1 element - <font color="red">NOT OK</font>std::copy_backward( example.begin()     , example.end() - 1 , example.end()       ); //shifts everything right 1 element - <font color="green">OK</font>std::copy_backward( example.begin() + 1 , example.end()     , example.end() - 1   ); //shifts everything left 1 element  - <font color="red">NOT OK</font>

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Originally posted by me22 Warning, I think you need to check that.