Type of pointer of array

Question

Below is the code

void printLoop(type?? p){

for(int i  = 0; i<2;i++)
{
   for(int e = 0;e<3;e++)
        {
             cout<<p[i][e]<<" ";
         }
      cout<<"\n";
  }
}
void array()
{
int a[2][3] = {{1,2,3},{4,5,6}};
int (*p)[3] = a;
printLoop(p);
}

Basic idea is that I want to print out the array using a for loop in the printLoop func. However, I need to know the type of that pointer which has the address of the 2D array. What's the pointer's type? Is it int (*)[]? I'm confused.

Also what does "(*p)" mean(from int (*p)[3]) ? Thanks a lot!

This is a really good read: c-faq.com/decl/spiral.anderson.html. From this, working counterclockwise you can see p is a pointer * to an array [] of 3 ints. — Qaz
– Qaz, Commented May 26, 2012 at 23:12

dirkgently · Accepted Answer · 2012-05-26 23:17:05Z

4

what does "(*p)" mean(from int (*p)[3]) ?

p is a pointer to an array of size 3 of objects of type int.

You have multiple possibilites for your printLoop function (though with the general C-restriction that you can leave at most one -- the outermost declarator empty):

You can specify the dimensions explicitly:

void printLoop(int p[ 2 ][ 3 ]);

The only advantage with this method is that the implementation can consider that the array being passed is of the desired size (i.e. 2x3 matrix of ints) as a pre-condition.

You can leave out the [ 2 ] part entirely:

void printLoop(int p[][ 3 ]);

or,

void printLoop(int (*p)[ 3 ]);

You can use a pointer to a pointer of int

You will also need to pass the dimensions (if you skip one that is) along to make sure that you don't access out-of-bounds memory. So, your function signature should go like this:

void printLoop(int (*p)[ 3 ], int dim);

answered May 26, 2012 at 23:17

dirkgently

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3 Comments

K-ballo Over a year ago

The first dimension will be lost either way, whether you define it or not. The array to pointer decay kicks in in function arguments as well.

dirkgently Over a year ago

@K-ballo: Yes. But the signature works as a sort of documentation for the implementer to take liberties.

whileone Over a year ago

Thanks so much. Now I have a better understanding of what's going on!

K-ballo · Accepted Answer · 2012-05-26 23:12:10Z

3

For the printLoop function, int p[2][3] as an argument should just work.

int (*p)[3] = a;

p is a pointer to an array of 3 ints, initialized to point to a.

answered May 26, 2012 at 23:12

K-ballo

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Comments

ScarletAmaranth · Accepted Answer · 2012-05-26 23:44:36Z

First of all, your code is not very modern C++. It's basically "c with iostreams".

Second of all, printLoop(int p[2][3]) is the signature you're looking for even though again, it's not the best way of doing things at all.

Third of all, int (*p)[3] is analyzed as follows: Start at the name which is p and take a look around (first to the right and then to the left yet here it doesn't matter) until you "hit" braces. There's only a star at it, so you can say that p is a pointer. Now you recursively do the same analysis again, you see [3], which means that p is a pointer to an array that has 3 ints.

Now I'd like to mention the following:

Use std::array for staticly-sized arrays.
Use std::vector for dynamicaly-sized arrays.

Oh, also, I myself wouldn't use a 2D array, they are clunky and just a syntactic sugar (around the basic "array" notion which is a syntactic sugar as well).

So perhaps, something like this, brain compiled, hopefully correct, C++11 abusing:

std::array<int, 3 * 2> p = {{1, 2, 3, 4, 5, 6}};
std::for_each(std::begin(p), std::end(p), [](int elem){ std::cout<<elem; });

Nice and dandy. You could also have lambda check for some "2d array" sizes and insert newlines if you so desire.

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