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I have a variable definition int (**ff)[4];, which is very bad looking. If I'm right (inferred from the fact that int (*f)[4]; is a pointer to an array of 4 int-s) this is a pointer to a pointer to an array of 4 int-s.

Now I have tried to initialize this thing but I had no success. Initializing f was simple (f = new int[5][4];), but the line ff = new int*[6][4]; is not working. Microsoft Visual Studio Community 2013 says in an error message that

a value of type "int *(*)[4]" cannot be assigned to an entity of type "int (**)[4]"

I have a very bad feeling that I really misunderstood something.

EDIT:

Sorry, that I didn't said it explicitly: What I wanted, is to allocate memory space for not only one, but for more pointers which later can point to an array of an array of 4 int-s. Of course I wanted this in one line without the help of any other definition, conversion etc.

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  • 3
    You see this variable, the first thing that comes to your mind is how to initialize it? I'd dump it :) Commented Feb 16, 2016 at 22:48
  • 1
    I just experimented with the language and wanted to understand better this part of it. Commented Feb 16, 2016 at 22:49
  • See std::array, std::vector. Commented Feb 16, 2016 at 22:50
  • @Bartis Point taken. Commented Feb 16, 2016 at 22:50
  • There are hundreds of different ways to properly initialize int (**ff)[4]. But there's no meaningless answer to your question until you explain what you want to use that int (**ff)[4] for. What are you trying to do? For example, you can initialize that ff with your own &f: int (**ff)[4] = &f;. Done. Does this help you? No one knows.... Commented Feb 16, 2016 at 23:38

4 Answers 4

Reset to default
4

I can (not) see this used only in this way:

int (*a)[4] = new int[6][4];
int (**ff)[4] = &a;

int (**ff)[4] means "a pointer to pointer to an array of four ints". This means we have to have int (*)[4] first - an lvalue, so we can take its address with operator&.

On the other hand, new int*[6][4]; allocates an array of pointers to an array of four pointers (the "outer" array decays). This is completely different type from the first one.

I had to help myself with cdecl.org on this one.

Edit:

I've just made this:

using PTR = int (*)[4];
int (**ff)[4] = new PTR; // hooray, a dynamically allocated pointer!

but can't figure a way without the type alias...

Actually, there is one: int (**ff)[4] = new (int (*)[4]), gathered from this Q&A. Don't forget that all you have now is a dynamically allocated uninitialized pointer.

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4 Comments

Another plan might be typedef int A4[4]; int (**ff)[4] = new A4*[length]; and the a loop, so you allocate a 3-d array with the first 2 dimensions variable and the third dimension 4.
@M.M I need a help out here! Can you figure out a way without a typedef? I've bumped into a syntax barrier.
You're not the first to wonder about this
I forgot about the extra-parentheses version from my own question, doh
0

A double pointer is a pointer to a pointer to get the obvious out of the way. That means that it will hold the memory address of a pointer, again obvious. That means you initialise your pointer and only then point your double pointer to it. Such as,

int* f[] = new int[4];
int** ff = &f;

Then to access this you can do,

(*ff)[1]

This will allow you to access the information stored in the dynamic array.

NOTE: leave the [] beside f empty if the size is not known at compile time which is the reason you should be doing it this way in the first place.

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0

If you want to access elements of an array a[3][3] by pointer to pointer to array than you can do that by below code : 

int a[3][3]={{1,2,3},{4,5,6},{7,8,9}};
int (*p)[3]=a;
int (**ff)[3]=&p;

To print it :

int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
printf("%d ",ff[0][i][j]);
}
printf("\n"); 
}
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-3

Alan,

I think the problem here is your initial assumption is incorrect. *f[4] (or (*f)[4]) would actually be an array of 4 pointers to int, not a pointer to an array of 4 ints. Then, **ff[4] would be a pointer to an array of 4 pointers to int. I would recommend backing up a little, with this information in hand, and trying it again. Also, an assignment to the **ff would be a single pointer, and probably not an array, depending on how you are intending to use it.

                            Blake Mitchell
                            retired free lance programmer
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2 Comments

int* f[4] is an array of 4 pointers to int, int (*f)[4] is a pointer to array of 4 ints.
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