Is there a more concise way to initialize a unique_ptr<char[]> than this?

Question

So at the moment I have:

std::string a;
std::unique_ptr<char[]> b(std::make_unique<char[]>(a.size() + 1));
std::copy(std::begin(a), std::end(a), b.get());

Is it is possible to initialize this directly in one step?

Answer 1

Is it is possible to initialize this directly in one step?

I would suggest keeping it as std::string or std::vector<char>.

However, if you really insist, Yes! Using an immediately invoking a lambda, this can be done.

std::unique_ptr<char[]> b = [&a]() {
   auto temp(std::make_unique<char[]>(a.size() + 1));
   std::copy(std::begin(a), std::end(a), temp.get());
   return temp;
}(); // invoke the lambda here!

The temp will be move constructed to the b.

(See a Demo)

---

If the string a will not be used later, you could move it to the std::unique_ptr<char[]>, using std::make_move_iterator.

#include <iterator>  // std::make_move_iterator

std::unique_ptr<char[]> b(std::make_unique<char[]>(a.size() + 1));
std::copy(std::make_move_iterator(std::begin(a)),
   std::make_move_iterator(std::end(a)), b.get());

If that needs to be in one step, pack it to the lambda like above.