New'ing a pointer to an array [Solved]

[Skippums]

Without using typedef, what is the syntax for dynamically allocating a pointer to an array? I tried the following, but can't get it to work:

long (*const val)[16] = new long()[16]; // C2440: Cannot convert from long* to long(*const)[16]
long (*const val)[16] = new long[16](); // This creates a (long*) with 16 elements

This is a question for the sake of curiosity, as I am actually using a typedef in the code to make it more clear. Thanks, EDIT: I actually can't figure out how to do this even when using typedefs :(. I have tried the following:

typedef long *myType[16]; // myType is an array of 16 pointers to long
myType *val = new myType(); // C2440: Cannot convert from long** to myType(*)

Any help with or without typedef's is appreciated. Thanks,

Sounds like somebody's got a case of the Mondays -Jeff

modified on Monday, November 29, 2010 3:27 PM

[_Superman_]

A pointer to an array is simple -

long* ptr = new long[16];

What you've done above is to create an array of pointers. So you actually have 16 pointers.

«_Superman_»  _I love work. It gives me something to do between weekends.

_Microsoft MVP (Visual C++)

Polymorphism in C

[Skippums]

I simplified my example. I am actually trying to dynamically create a 2D array, where one dimension is known (the unknown dimension needs to be consecutive in memory), hence I have:

long *var[16];

However, I would like to create a value for "var" in a function, and return it. Hence I have:

long *(&myFunc())[16] {
long *(*rval)[16] = new ?????;
...
return *rval;
}

Then I declare my variable as a reference and delete it at the end of the code

long *(&var)[16] = myFunc();
...
delete &var;

Is there some better way to go about this? I could just make it doubly-dynamic, but feel like I should retain known dimensions for the sake of readability. Any ideas? Thanks,

Sounds like somebody's got a case of the Mondays -Jeff

[_Superman_]

Here is how it can be done, but its real ugly -

#define FIXED_SIZE 16

void allocate(long (**ptr)[FIXED_SIZE])
{
*ptr = new long[10][FIXED_SIZE];
}

int _tmain(int argc, _TCHAR* argv[])
{
long (*ptr)[FIXED_SIZE];

allocate(&ptr);

delete \[\] ptr;

return 0;

}

The clean and recommended way is to use a standard container - std::vector<std::vector<long>> longArr;

«_Superman_»  _I love work. It gives me something to do between weekends.

_Microsoft MVP (Visual C++)

Polymorphism in C

modified on Monday, November 29, 2010 2:46 PM

[Skippums]

I guess the answer is, "you can't allocate a single pointer to an array without the new[] operator" (with brackets). So, to allocate a single pointer to an array you have to do:

long (*val)[16] = new long[1][16];
...
delete[] val;

Thanks,

Sounds like somebody's got a case of the Mondays -Jeff

[_Superman_]

This is a single pointer to an array -

long* ptr = new long[16];
delete [] ptr;

Now I'm not sure what you're looking for.

«_Superman_»  _I love work. It gives me something to do between weekends.

_Microsoft MVP (Visual C++)

Polymorphism in C

[Skippums]

Your last post solved it... I just can't allocate a SINGLE pointer to an array of pointers without dynamically allocating an array of 1 pointer to the array. For example:

// Allocate a single pointer to an array of 16 pointers to long
// Can't be done without using operator new[]
long *(*val)[16] = new ???;
...
delete val;

There is nothing I can put in the "???" area to make the above code valid, because there is no valid syntax to dynamically create a single pointer to array using operator new(). I must do the following instead:

// Allocate a single pointer to an array of 16 pointers to long
long *(*val)[16] = new long*[1][16];
...
delete[] val;

In short, you have answered my question. Thanks,

Sounds like somebody's got a case of the Mondays -Jeff

[_Superman_]

:thumbsup:

«_Superman_»  _I love work. It gives me something to do between weekends.

_Microsoft MVP (Visual C++)

Polymorphism in C