I'm having a brain fart at the moment and I am looking for a fast way to take an array and pass half of it to a function. If I had an array A of ten elements, in some languages I could pass something like A[5:] to the function and be done with it. Is there a similar construct in c++? Obviously I'd like to avoid and sort of looping function.
2 Answers
Yes. In plain C you use pointers, but in C++ you can use any kind of iterator (a pointer can be considered an iterator).
template<typename Iter>
void func(Iter arr, size_t len) { ... }
int main() {
int arr[10];
func(arr, 10); // whole array
func(arr, 5); // first five elements
func(arr + 5, 5); // last five elements
std::vector<Thing> vec = ...;
func(vec.begin(), vec.size()); // All elements
func(vec.begin(), 5); // first five
func(vec.begin() + 5, vec.size() - 5); // all but first 5
return 0;
}
The typical trick is to pass a pointer to the first element of the array, and then use a separate argument to pass the length of the array. Unfortunately there are no bounds checks, so you have to be careful to get it right or you will scribble on your memory.
You can also use half-open ranges. This is the most common way to do it. Many functions in the standard library (like std::sort) work this way.
template<class Iter>
void func(Iter start, Iter end) { ... }
int main() {
int arr[10];
func(arr, arr + 10); // whole array
func(arr, arr + 5); // first five elements
func(arr + 5, arr + 10); // last five elements
std::vector<Thing> vec = ...;
func(vec.begin(), vec.end()); // whole vector
func(vec.begin(), vec.begin() + 5); // first five elements
func(vec.begin() + 5, vec.end()); // all but the first five elements
return 0;
}
Again, no bounds checks.
4 Comments
Corbin
I think the templating you were talking about is:
template<Iter> void func(const Iter& start, const Iter& end) { ... }
K-ballo
Just
template< typename Iter > void func( Iter start, Iter end )
jthill
if you want bounds checks use bounds-checked access,
&vec.at(5), or vec.begin()+max(5,vec.size()) if '5' is really 'at most 5'.
Dietrich Epp
@jthill: The "max" solution can be problematic because it silently behaves differently, rather than throwing an error. I would usually rather have the error. The "at" solution doesn't give you an iterator, which means you can't pair it with
vec.end() but you have to do something like &*vec.end() instead.I also had the same use but instead I used vector and used the syntax
std::vector <int> a(10);
// for example to use by removing first element
a = std::vector<int>(a.begin() + 1, a.end())
//its ur turn to change the size
1 Comment
qwr
This vector constructor copies all elements, correct?
std::vector, which has a dual-iterator constructor:std::vector<int> a{1,2,3,4,5,6,7,8,9,10}; foo(std::vector<int>(std::next(std::begin(a), a.size() / 2), std::end(a)));This strategy really pales in comparison to iterators in terms of speed, however, not to mention easy use.