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1234567891011121314 | #include <vector>
#include <algorithm>
int main()
{
std::vector<int> v = {1, 2, 3, 4, 2, 5, 2, 6};
v.erase(std::remove(std::begin(v), std::end(v), 2),
std::end(v));
v.erase(std::remove_if(std::begin(v), std::end(v),
[](int i) { return i%2 == 0; }),
std::end(v));
} |
This pattern is licensed under the CC0 Public Domain Dedication.
Use the erase-remove idiom to remove elements from a container.
On line 6, we create a std::vector as an example
container and initialize it some int elements.
The std::remove and std::remove_if
algorithms do not have knowledge about the underlying storage of
the given range, so cannot actually remove elements from that
storage. Instead, these algorithms actually shift (by means of move
assignment) the elements in the range in such a way that the
elements not removed form a new range at the beginning of the
original range. The algorithms return a past-the-end iterator for
this new range, which, since the removed elements have been shifted
to the end of the original range, also marks the start of the
removed elements.
On lines 89, we show how we can pass the resulting iterator to the
containers erase member function,
which actually removes the elements from the container. Similarly,
on lines 1113, we demonstrate how std::remove_if can be used with
erase to remove all elements for which a given predicate returns
true (in the example code, we remove all even elements).
This technique of using the generic remove algorithms followed by
a call to the particular containers erase member function is
commonly referred to as the erase-remove idiom.
10 December 2017
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