Multi-Dim Arrays

[Suko]

Hi,

Is this is a 2D or 3D array ?
int (*ptr)[10][15] = new int[value][10][15];

If its 2D what affect is 'value' having ?
Accessing the data is don by (*ptr)[0][0] = 11; which says its 2D, but you can use ptr[0][0][0] = 11; which makes it look 3D!

So if value was 3, what have i created ?

[Savage]

Hi,

Is this is a 2D or 3D array ?
int (*ptr)[10][15] = new int[value][10][15];

If its 2D what affect is 'value' having ?
Accessing the data is don by (*ptr)[0][0] = 11; which says its 2D, but you can use ptr[0][0][0] = 11; which makes it look 3D!

So if value was 3, what have i created ?

It's a pointer to 2D array, or with other words, you can call it 3D array.

Savage

[weaknessforcats]

Is this is a 2D or 3D array ?
int (*ptr)[10][15] = new int[value][10][15];

If its 2D what affect is 'value' having ?

Without blowing your mind, it's a one-dimensional array.

First, there are only one-dimensional arrays in C or C++. The number of elements in put between brackets:

int array[5];
 

That is an array of 5 elements each of which is an int.

int array[];
 

won't compile. You need to declare the number of elements.

Second, this array:

int array[5][10];
 

is still an array of 5 elements. Each element is an array of 10 int.

int array[5][10][15];
 

is still an array of 5 elements. Each element is an array of 10 elements where each element is an array of 15 int.

int array[][10];
 

won't compile. You need to declare the number of elements.

Third, the name of an array is the address of element 0

int array[5];
 

Here array is the address of array[0]. Since array[0] is an int, array is the address of an int. You can assign the name array to an int*.

int array[5][10];
 

Here array is the address of array[0]. Since array[0] is an array of 10 int, array is the address of an array of 10 int. You can assign the name array to a pointer to an array of 10 int:

int array[5][10];
 
int (*ptr)[10] = array;
 

Fourth, when the number of elements is not known at compile time, you create the array dynamically:

int* array = new int[value][5];
int (*ptr)[10] = new int[value][10];
int (*ptr)[10][15] = new int[value][10][15];
 

In each case value is the number of elements. Any other brackets only describe the elements.

Using an int** for an array of arrays is incorrect and produces wrong answers using pointer arithmetic. The compiler knows this so it won't compile this code:

int** ptr = new int[value][10];    //ERROR
 

new returns the address of an array of 10 int and that isn't the same as an int**.

Likewise:

int*** ptr = new int[value][10][15];    //ERROR
 

new returns the address of an array of 10 elements where each element is an array of 15 int and that isn't the same as an int***.