Problems with a Dynamic Array of Static Arrays

Question

I do not plan on changing the size of my dynamic array. The reason I want to create a dynamic array of static arrays (which contain shorts) is so that I can return that array in a function (not defining the size of that array until I'm in that function).

My first question then, is how do I return such an array? Functions don't let you return a pointer without a type & I'm unable to find out how to define a pointer with type array.

This brings me to my second question, which is how do I correctly define a dynamic array of static arrays? I have searched for this online, but none of the answers have been too helpful.

One way to do it is to declare: short (*array)[size] but the problem with this is that I don't know how to initialize the array in that case, and size has to be a literal.

I could do this:

typedef short column[size];
column * row = NULL;
row = malloc(rowMax * sizeof(column));
row[0][0] = 10;

but again, size has to be a literal; and even if size is a literal, I receive an error stating that a value of type "void *" cannot be assigned to an entity of type "column *".

If any of you have a solution that doesn't use vectors that would be greatly appreciated; as the vector class is larger than the array class.

"as the vector class is larger than the array class." There's no array class in your sample. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Mar 3, 2016 at 23:54
And if you are using C++11 or later, you can use std::array for the static array. — Remy Lebeau
– Remy Lebeau, Commented Mar 3, 2016 at 23:54
@RemyLebeau I tried using: short (*array)[size] = new std::array[row]; but that results in the error: argument list for class template std::array is missing. — Larrimus
– Larrimus, Commented Mar 3, 2016 at 23:58
What is your version of C++? Please show enough code to generate your errors. Please edit your updates into your question. — philipxy
– philipxy, Commented Mar 4, 2016 at 0:03

jotik · Accepted Answer · 2016-03-04 00:02:12Z

2

In C++11 one could use std::array from <array> and do:

constexpr std::size_t const cols = 42u; // static size
using MyStaticArray = std::array<short, cols>; // static array type
// Or in C++03 and earlier: typedef short MyStaticArray[42u];

std::size_t const rows = someValue; // dynamic size
MyStaticArray * dynamicArray = new MyStaticArray[rows];
// Or better: std::vector<MyStaticArray> dynamicArray(rows);

for (std::size_t row = 0u; row < rows; ++row)
  for (std::size_t col = 0u; col < cols; ++col)
    dynamicArray[row][col] = rows * cols;

PS: Also give upvotes to Remy Lebeau below for his comments helped to improve this post.

edited Mar 4, 2016 at 0:02

answered Mar 3, 2016 at 23:58

jotik

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1 Comment

Remy Lebeau Over a year ago

If you are going to go to the trouble of using std::array, you may as well use std::vector for the dynamic array: std::vector<MyStaticArray> dynamicArray(rows); Even if you are not using C++11, you can still use std::vector, just typedef the static array using typedef short MyStaticArray[cols];

bleepbloop · Accepted Answer · 2016-03-04 00:05:23Z

1

For the sake of saving time, just create a vector of static arrays

std::vector<std::array<Type, n>>

Not sure why the size difference matters, it's only a few Kb.

Also, if you're going to write C code in a C++ app... you're gunna have a bad time.

answered Mar 4, 2016 at 0:05

bleepbloop

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