How to Initialise a Dynamic Array in C++

How do you initialise a dynamic array in C++?

char* c = new char[length]();

Initialize values of dynamic array in a function (C language)

You have the address of a pointer passed to the function:

array

You want to dereference that to get the pointer:

*array

Then apply the array subscript operator to the result:

(*array)[2]

Or equivalently:

*((*array) + 2)

The parenthesis are required because the array subscript operator [] has higher precedence than the dereference operator *.

Generally speaking, you should use the array subscript operator whenever possible, as it tends to be easier to read.

Dynamic array initialization in c

can we do this?

No. But you can do this:

int arr[n];
memset(arr, 0, sizeof(arr));

You lose the syntactic sugar for initialization, but you get the functionality.

Initialize a dynamic array in O(1) time in C/C++

No, there is not, not in c++, nor in c, not in any language or CPU I know of. But it can be done with 1 line of code.

in C: for a array of char:

char a[N];
memset(a, '1', N);  // only works for data 1 byte long

in C++

std::vector<T> v;
std::fill(v.begin(), v.end(), value);

How to initialize a dynamic int array elements to 0 in C

In this case you would use calloc():

array = (int*) calloc(n, sizeof(int));

~~It's safe to assume that all systems now have all zero bits as the representation for zero.~~

§6.2.6.2 guarantees this to work:

For any integer type, the object representation where all the bits are
zero shall be a representation of the value zero in that type.

It's also possible to do a combination of malloc() + memset(), but for reasons discussed in the comments of this answer, it is likely to be more efficient to use calloc().

Initialize a dynamic C array within a C++ initializer list

I don't understand how the C array is being initialized within the initializer list.

here:

myArray(int size) : size(size), elements(new int[size])

elements are basicly assigned value returned by new int[size]

You could rewrite it as :

myArray(int size)
{
    this->size = size;
    elements = new int[size];
    std::cout << "Constructed array of size " << size << std::endl;
}

[edit]

This:

// elements(new int[size]); // why will line only work in initializer list?

will not work in the function body, because in this place compiler thinks that you want to call a function named elements, but there is no such function.

The fact that it works in the initializer list is because that is how the language syntax was made.

Seg faulting with 4D arrays & initializing dynamic arrays

Even though I have not understood why do you use 4D arrays to store shapes for a tetris game, and I agree with bolov's comment that such an array should not overflow the stack (7*4*4*4*1 = 448 bytes), so you should probably check other code you wrote.

Now, to your question on how to manage 4D (N-Dimensional)dynamically sized arrays. You can do this in two ways:

The first way consists in creating an array of (N-1)-Dimensional arrays. If N = 2 (a table) you end up with a "linearized" version of the table (a normal array) which dimension is equal to R * C where R is the number of rows and C the number of columns. Inductively speaking, you can do the very same thing for N-Dimensional arrays without too much effort. This method has some drawbacks though:
- You need to know beforehand all the dimensions except one (the "latest") and all the dimensions are fixed. Back to the N = 2 example: if you use this method on a table of C columns and R rows, you can change the number of rows by allocating C * sizeof(<your_array_type>) more bytes at the end of the preallocated space, but not the number of columns (not without rebuilding the entire linearized array). Moreover, different rows must have the same number of columns C (you cannot have a 2D array that looks like a triangle when drawn on paper, just to get things clear).
- You need to carefully manage the indicies: you cannot simply write my_array[row][column], instead you must access that array with my_array[row*C + column]. If N is not 2, then this formula gets... interesting
You can use N-1 arrays of pointers. That's my favourite solution because it does not have any of the drawbacks from the previous solution, although you need to manage pointers to pointers to pointers to .... to pointers to a type (but that's what you do when you access my_array[7][4][4][4].

Solution 1

Let's say you want to build an N-Dimensional array in C using the first solution.
You know the length of each dimension of the array up to the (N-1)-th (let's call them d_1, d_2, ..., d_(N-1)). We can build this inductively:

We know how to build a dynamic 1-dimensional array
Supposing we know how to build a (N-1)-dimensional array, we show that we can build a N-Dimensional array by putting each (N-1)-dimensional array we have available in a 1-Dimensional array, thus increasing the available dimensions by 1.

Let's also assume that the data type that the arrays must hold is called T.
Let's suppose we want to create an array with R (N-1)-dimensional arrays inside it. For that we need to know the size of each (N-1)-dimensional array, so we need to calculate it.

For N = 1 the size is just sizeof(T)
For N = 2 the size is d_1 * sizeof(T)
For N = 3 the size is d_2 * d_1 * sizeof(T)

You can easily inductively prove that the number of bytes required to store R (N-1)-dimensional arrays is R*(d_1 * d_2 * ... * d_(n-1) * sizeof(T)). And that's done.

Now, we need to access a random element inside this massive N-dimensional array. Let's say we want to access the item with indicies (i_1, i_2, ..., i_N). For this we are going to repeat the inductive reasoning:

For N = 1, the index of the i_1 element is just my_array[i_1]
For N = 2, the index of the (i_1, i_2) element can be calculated by thinking that each d_1 elements, a new array begins, so the element is my_array[i_1 * d_1 + i_2].
For N = 3, we can repeat the same process and end up having the element my_array[d_2 * ((i_1 * d_1) + i_2) + i_3]

And so on.

Solution 2

The second solution wastes a bit more memory, but it's more straightforward, both to understand and to implement.

Let's just stick to the N = 2 case so that we can think better. Imagine to have a table and to split it row by row and to place each row in its own memory slot. Now, a row is a 1-dimensional array, and to make a 2-dimensional array we only need to be able to have an ordered array with references to each row. Something like the following drawing shows (the last row is the R-th row):

+------+
|  R1 -------> [1,2,3,4]
|------|
|  R2 -------> [2,4,6,8]
|------|
|  R3 -------> [3,6,9,12]
|------|
| .... |
|------|
|  RR -------> [R, 2*R, 3*R, 4*R]
+------+

In order to do that, you need to first allocate the references array (R elements long) and then, iterate through this array and assign to each entry the pointer to a newly allocated memory area of size d_1.

We can easily extend this for N dimensions. Simply build a R dimensional array and, for each entry in this array, allocate a new 1-Dimensional array of size d_(N-1) and do the same for the newly created array until you get to the array with size d_1.

Notice how you can easily access each element by simply using the expression my_array[i_1][i_2][i_3]...[i_N].

For example, let's suppose N = 3 and T is uint8_t and that d_1, d_2 and d_3 are known (and not uninitialized) in the following code:

size_t d1 = 5, d2 = 7, d3 = 3;
int ***my_array;

my_array = malloc(d1 * sizeof(int**));
for(size_t x = 0; x<d1; x++){
    my_array[x] = malloc(d2 * sizeof(int*));
    for (size_t y = 0; y < d2; y++){
         my_array[x][y] = malloc(d3 * sizeof(int));
    }
}

//Accessing a random element
size_t x1 = 2, y1 = 6, z1 = 1;    
my_array[x1][y1][z1] = 32;

I hope this helps. Please feel free to comment if you have questions.

Initializing dynamic array of unions in C

To define a union type do

union element
{
    int digit;
    char letter;
    float number;
};

To dynamical allocated an array with n elements of type union element do:

union element * pu = malloc(n * sizeof (union element));

or even better

union element * pu = malloc(n * sizeof *pu);

The code you show here

union element
{
    int digit;
    char letter;
    float number;
} f;

does two things:

Define a type union element
Define the variable f to be of type union element

Note

If this

 element* arr = malloc(n * ...

compiles, then you are not using a C compiler, but a C++ compiler.

A C compiler required

 union element* arr = malloc(n * ...

A final comment on your wording: "Initiallizing ... array ...":

The code dynamical creates n elements of a union. Those element do not get "initialized".

To have them being initialised during allocation of calloc() instrad of malloc();

... = calloc(n, sizeof ...); /* Note the different number of parameters. */

C allocating dynamic array of doubles and initializing it with memset

because memset works with bytes

sizeof(double)==8,
so each double in your array is filled with the value 0x0101010101010101

just replace the memset with:

for (int i=0;i< numElements;i++) d[i]=1;