dynamic memory?

why is this not possible?

im trying to declare an two dimentional array this way but i get an error message "Cannot convert 'int ( *)[2]' to 'int *'

Code:

#include <iostream.h> int main()  {int * number; number = new int [5][2]; return 0; }

sorry i cant use insert source code feature well

I think the probelm is you have not declared number as an 2D array and possibly you have not declared new int as well.

Code:

int (* number)[2];

for multi dimensional arrays the compiler associates the array elements as part of the type so the pointer type has to match.

int array[5][2] is of the type int[5][2] so the pointer has to be of type int *[2]. You need the parenthesis otherwise it will declare an array of pointers.

If you want a set size to your array then declare it as an array of pointers and dynamically allocate memory to each pointer:

Code:

int *array[5];
for ( int i ( 0 ); i < 5; i++ )
  array[i] = new int[2];

If the first dimension needs to be variable as well as subsequent dimensions then you will need to declare the array as a pointer to a pointer and allocate memory on both ends:

Code:

int **array = new int*[5];
for ( int i ( 0 ); i < 5; i++ )
  array[i] = new int[2];

With this second method, remember that the levels of indirection match the number of dimensions. And don't forget to release the memory.

Code:

for ( int i ( 0 ); i < 5; i++ )
  delete [] array[i];
delete [] array;

hmm.... it's hard to undestand what you guys are saying but i think i get the idea. thanks for the help, and by the way, why do i have to release the memory? do i relelease it at the end of the program or at the moment i want it to be released?

You need to release the memory, otherwise you have a 'memory leak', which will eventually lead to your computer crashing if you run the program too many times You can release it any time when you're done with it. Hopefully, that will be at some time before the program ends, otherwise you have some problems

ooooooo ok. i just came up with another question. how do you pass an array from function to main function? oh and when using dynamic memory, does pointer become an array or just act as if it was an array? is 2d array an array or a pointer?

>how do you pass an array from function to main function?
It depends on how you declared the array. Just copy the declaration into your function parameter list and that should be enough.

>does pointer become an array or just act as if it was an array?
Nice catch. Not many people understand this. No, it simulates an array by taking advantage of the fact that an array name is often converted to a pointer to the first element and array subscripting is really pointer arithmetic. A dynamic array is not an array, even though it pretends to be one pretty well.

>is 2d array an array or a pointer?
If your declaration is an array, then it's an array. But beware the sneaky parentheses, they may try to trick you:

Code:

int a[10]; // Array
int b[10][10]; // Array
int *c[10]; // Array
int *d; // Pointer
int **e; // Pointer
int (*f)[10]; // Pointer

ok thanks prelude, you sure do know a lot of things about programming

Code:

int a[10]; // Array
int b[10][10]; // Array
int *c[10]; // Array
int *d; // Pointer
int **e; // Pointer
int (*f)[10]; // Pointer

That's why I prefer to stick to 1-dimensional arrays