Copy section of an array into a std::vector in C++

Question

Hello im wondering if its possible to copy a section of an array whilst initializing a vector.

For example I have tried

UINT16 myarray[] = { 0,1,2,3,4,5,6, ..., n };
std::vector<UINT16> myvect(myarray[3], myarray[3] + 16);

This kind of works however I seem to get more data than i intended. Is this the correct usage?

You need to pass addresses, so `myarray + 3, myarray + 19` seems like it would work. Better to identifier your pointers though. — François Andrieux, Jan 08 '19 at 20:10
Or take the addresses like this: `std::vector myvect(&myarray[3], &myarray[3 + 16])` — Galik, Jan 08 '19 at 20:37
As François Andrieux and Jarod42 stated, it was a matter of using integer offsets to select which part of the array i wanted copied. It worked out, thanks! — Chris, Jan 08 '19 at 23:12

score 1 · Accepted Answer · answered Jan 08 '19 at 20:36

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The way is to pass iterators, not values

std::vector<UINT16> myvect(myarray + 3, myarray + 3 + 16);

answered Jan 08 '19 at 20:36

Jarod42

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Assuming size is greater than 18, else you have UB. – Sid S Jan 08 '19 at 22:00
@SidS : :-) :-) – Jarod42 Jan 08 '19 at 22:02

Sid S · Answer 2 · 2019-01-08T21:41:15.107

1

UINT16 myarray[] = { 0,1,2,3,4,5,6, ..., n };
std::vector<UINT16> myvect(myarray[3], myarray[3] + 16);

This uses myarray[3] which is 3 and myarray[3] + 16 which is 19 as arguments to the myvect constructor. Which in turn sets up myvect to contain 3 elements, all with the value 19.

If you want a copy of the 16 elements starting at index 3, you can do it like this:

std::vector<UINT16> myvect(&myarray[3], &myarray[3 + 16]);

edited Jan 08 '19 at 21:41

answered Jan 08 '19 at 20:43

Sid S

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Assuming size is greater than 19, else you have UB. (Where it is clear that `UINT16 myarray[7]` would produce UB, it is also the case for `UINT16 myarray[19]`). – Jarod42 Jan 08 '19 at 20:52
@Jarod42, It will work if size is 19, too. Anyway, this possible UB was present in the question and it's not what the OP was asking about. – Sid S Jan 08 '19 at 21:44
Seems there is a debate if `&*pointer` is `pointer` or not (and so UB or not)... So I would avoid it. See [may-i-take-the-address-of-the-one-past-the-end-element-of-an-array](https://stackoverflow.com/questions/3144904/may-i-take-the-address-of-the-one-past-the-end-element-of-an-array). – Jarod42 Jan 08 '19 at 21:52

Robert Andrzejuk · Answer 3 · 2019-01-08T23:18:04.567

Here is a different (more verbose) way:

uint16_t       myarray[]  = { 0, 1, 2, 3, 4, 5, 6, ..., n };
constexpr auto arr_size   = sizeof(myarray) / sizeof(myarray[0]);
constexpr auto copy_start = std::min< size_t >(3, arr_size);
constexpr auto copy_count = 19;
constexpr auto copy_end   = std::min< size_t >(copy_start + copy_count, arr_size);

vector< uint16_t > myvect(std::next(myarray, copy_start), std::next(myarray, copy_end));

Instead of UINT16 you can use uint16_t type from C++11 onwards: Fixed width integers (if your compiler supports it)

If your values are available at compile time, then you can use compile time constants(constexpr) to calculate the helper variables. Otherwise const should be used.

Standard algorithms can be used also on arrays so providing an offset to std::next can be used to get the n-th element of an array.

Copy section of an array into a std::vector in C++

3 Answers3