Initializer list for dynamic arrays?

Question

It is possible to give an initializer list to the definition of a static array. Example:

int main()
{
  int int_static[2] = {1,2};
}

Is a s

Answer 1

At the time the OP posted this question, C++11 support may not have been very prevalent yet, which is why the accepted answer says this is not possible. However, initializing a dynamic array with an explicit initializer list should now be supported in all major C++ compilers.

The syntax new int[3] {1, 2, 3} was standardized in C++11. Quoting the new expression page on cppreference.com:

The object created by a new-expression is initialized according to the following rules:
...
If type is an array type, an array of objects is initialized:
...
If initializer is a brace-enclosed list of arguments, the array is aggregate-initialized. (since C++11)

So, given the OP's example, the following is perfectly legal when using C++11 or newer:

foo * foo_array = new foo[2] { nullptr, nullptr };

Note that by providing pointers in the initializer list, we're actually coaxing the compiler to apply the foo(void * ptr) constructor (rather than the default constructor), which was the desired behavior.

Answer 2

Given that you're real class is more complex than an int, and constructed from differing values, it's complicated. A vector can be constructed with iterators if you have an existing array/vector with the correct values to default from, or you have to use placement new.

//vector
int main()
{
  int int_static[2] = {1,2};
  std::vector<int> int_dynamic(int_static, int_static+2);
  //this is what everyone else is saying.  For good reason.
}
//placement new
int function_that_returns_constructed_from_values() {
    return rand();
}
int main() 
{
    int count = 2;
    char *char_dynamic = new char[count * sizeof(int)];
    int *int_dynamic = char_dynamic;
    for(int i=0; i<count; ++i)
        new(int_dynamic+i)int(function_that_returns_constructed_from_values());
    //stuff
    for(int i=0; i<count; ++i)
        (int_dynamic+i)->~int(); //obviously not really int
    delete []char_dynamic;
}

Obviously, the vector is the preferred way to do this.

Answer 3

The initializer data must be somewhere anyway. Simply name it.

E.g.,

#include <stddef.h>
#include <algorithm>        // std::copy
#include <vector>

typedef ptrdiff_t   Size;

template< class Type, Size n >
Size countOf( Type (&)[n] ) { return n; }

int main()
{
    using namespace std;

    static int const    initData[]  = {1,2};
    static Size const   n           = countOf( initData );

    // Initialization of a dynamically allocated array:
    int*        pArray  = new int[n];
    copy( initData, initData + n, pArray );

    // Initialization of a vector:
    vector<int> v( initData, initData + n );
}

EDIT: fixed a thinko in above code. I hastened to add example on request. So what I put did erroneously use return value from std::copy.

Cheers & hth.,

Answer 4

No, you will have to create the elements dynamically.

Alternatively, you can use a local array and copy its elements over those of the dynamically allocated array:

int main() {
   int _detail[] = { 1, 2 };
   int * ptr = new int[2];
   std::copy( _detail, _detail+(sizeof detail / sizeof *detail), ptr );
   delete [] ptr;
}

In the limited version of setting all elements to 0, you can use an extra pair of parenthesis in the new call:

int * ptr = new int[2]();  // will value initialize all elements

But you seem to be looking for a different thing.

Answer 5

No, you cannot do that.

I think C++ doesn't allow this because allowing such thing doesn't add any nice-to-have feature to the language. In other words, what would be the point of dynamic array if you use a static initializer to initialize it?

The point of dynamic array is to create an array of size N which is known at runtime, depending on the actual need. That is, the code

int *p = new int[2]; 

makes less sense to me than the following:

int *p = new int[N]; //N is known at runtime

If that is so, then how can you provide the number of elements in the static initializer because N isn't known until runtime?

Lets assume that you're allowed to write this:

int *p = new int[2] {10,20}; //pretend this!

But what big advantage are you getting by writing this? Nothing. Its almost same as:

int a[] = {10,20};

The real advantage would be when you're allowed to write that for arrays of N elements. But then the problem is this:

 int *p = new int[N] {10,20, ... /*Oops, no idea how far we can go? N is not known!*/ };