Say we have two arrays:
double *matrix=new double[100];
double *array=new double[10];
And we want to copy 10 elements from matrix[80:89] to array using memcpy.
Any quick solutions?
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1would that code compile? – Aaron Anodide Oct 10 '10 at 21:15
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1I think that's not even a valid c++ declaration of an array. memcpy requires a start and a length + size of the variable. The Parameters have to be something like matrix = src, array = dest, 80 = start, 10 = len, sizeof(double) = size . $ man memcopy – Ronny Brendel Oct 10 '10 at 21:16
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5Do you perhaps mean `double matrix[100];` and `double array[10];`? – James McNellis Oct 10 '10 at 21:16
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yeah...that code doesn't look right in any language.. – mpen Oct 10 '10 at 21:18
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@original poster: if you correct the sample code i'll bump the points back - the answers are useful enough; the question is misleading though (incorrect code) – Aaron Anodide Oct 10 '10 at 21:24
3 Answers
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It's simpler to use std::copy:
std::copy(matrix + 80, matrix + 90, array);
This is cleaner because you only have to specify the range of elements to be copied, not the number of bytes. In addition, it works for all types that can be copied, not just POD types.
James McNellis
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memcpy(array, &matrix[80], 10*sizeof(double));
But (since you say C++) you'll have better type safety using a C++ function rather than old C memcpy:
#include <algorithm>
std::copy(&matrix[80], &matrix[90], array);
Note that the function takes a pointer "one-past-the-end" of the range you want to use. Most STL functions work this way.
aschepler
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5One advantage of using the `matrix + N` instead of `&matrix[N]` is that for an array of size `M`, `matrix + M` is allowed but `&matrix[M]` is not (because the latter dereferences a pointer to the element one-past-the-end). – James McNellis Oct 10 '10 at 21:47
