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
– François Andrieux, Commented Jan 8, 2019 at 20:10
Or take the addresses like this: std::vector<UINT16> myvect(&myarray[3], &myarray[3 + 16]) — Galik
– Galik, Commented Jan 8, 2019 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
– Chris, Commented Jan 8, 2019 at 23:12

Jarod42 · Accepted Answer · 2019-01-08 20:36:44Z

1

The way is to pass iterators, not values

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

answered Jan 8, 2019 at 20:36

Jarod42

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2 Comments

Sid S Over a year ago

Assuming size is greater than 18, else you have UB.

Jarod42 Over a year ago

@SidS : :-) :-)

Sid S · Accepted Answer · 2019-01-08 21:41:15Z

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 8, 2019 at 21:41

answered Jan 8, 2019 at 20:43

Sid S

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3 Comments

Jarod42 Over a year ago

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]).

Sid S Over a year ago

@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.

Jarod42 Over a year ago

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.

Robert Andrzejuk · Accepted Answer · 2019-01-08 23:18:04Z

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.

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