std::copy, std::copy_if
Defined in header
<algorithm>
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template< class InputIt, class OutputIt >
OutputIt copy( InputIt first, InputIt last, OutputIt d_first ); |
(1) | |
template< class InputIt, class OutputIt, class UnaryPredicate >
OutputIt copy_if( InputIt first, InputIt last, |
(2) | (since C++11) |
Copies the elements in the range, defined by [first, last)
, to another range beginning at d_first
. The second function only copies the elements for which the predicate pred
returns true. The order of the elements that are not removed is preserved.
Contents |
[edit] Parameters
first, last | - | the range of elements to copy |
d_first | - | the beginning of the destination range. If d_first is within [first, last) , std::copy_backward must be used instead of std::copy.
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pred | - | unary predicate which returns true for the required elements. The signature of the predicate function should be equivalent to the following: bool pred(const Type &a); The signature does not need to have const &, but the function must not modify the objects passed to it. |
Type requirements | ||
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InputIt must meet the requirements of InputIterator .
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OutputIt must meet the requirements of OutputIterator .
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UnaryPredicate must meet the requirements of Predicate .
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[edit] Return value
Output iterator to the element in the destination range, one past the last element copied.
[edit] Complexity
1) Exactly last - first
assignments
2) Exactly last - first
applications of the predicate
[edit] Notes
In practice, implementations of std::copy
avoid multiple assignments and use bulk copy functions such as std::memmove if the value type is TriviallyCopyable
When copying overlapping ranges, std::copy
is appropriate when copying to the left (beginning of the destination range is outside the source range) while std::copy_backward
is appropriate when copying to the right (end of the destination range is outside the source range).
[edit] Possible implementation
First version |
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template<class InputIt, class OutputIt> OutputIt copy(InputIt first, InputIt last, OutputIt d_first) { while (first != last) { *d_first++ = *first++; } return d_first; } |
Second version |
template<class InputIt, class OutputIt, class UnaryPredicate> OutputIt copy_if(InputIt first, InputIt last, OutputIt d_first, UnaryPredicate pred) { while (first != last) { if (pred(*first)) *d_first++ = *first; first++; } return d_first; } |
If you do not have C++11, an equivalent to std::copy_if is to use std::remove_copy_if with the negated predicate.
template<class InputIt, class OutputIt, class UnaryPredicate> OutputIt copy_if(InputIt first, InputIt last, OutputIt d_first, UnaryPredicate pred) { return std::remove_copy_if(first, last, d_first, std::not1(pred)); } |
[edit] Example
The following code uses copy to both copy the contents of one vector to another and to display the resulting vector:
#include <algorithm> #include <iostream> #include <vector> #include <iterator> #include <numeric> int main() { std::vector<int> from_vector(10); std::iota(from_vector.begin(), from_vector.end(), 0); std::vector<int> to_vector; std::copy(from_vector.begin(), from_vector.end(), std::back_inserter(to_vector)); // or, alternatively, // std::vector<int> to_vector(from_vector.size()); // std::copy(from_vector.begin(), from_vector.end(), to_vector.begin()); // either way is equivalent to // std::vector<int> to_vector = from_vector; std::cout << "to_vector contains: "; std::copy(to_vector.begin(), to_vector.end(), std::ostream_iterator<int>(std::cout, " ")); std::cout << '\n'; }
Output:
to_vector contains: 0 1 2 3 4 5 6 7 8 9
[edit] See also
copies a range of elements in backwards order (function template) |
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(C++11)
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copies a number of elements to a new location (function template) |
copies a range of elements omitting those that satisfy specific criteria (function template) |