finding size of multidimensional array

Question

How would I find the size in bytes of an array like this?

    double dArray[2][5][3][4];

I've tried this method but it returns "2". Maybe because the array is only declared and not filled with anything?

     #include <iostream>
     #define ARRAY_SIZE(array) (sizeof((array))/sizeof((array[0])))
     using namespace std;

      int main() {
        double dArray[2][5][3][4];
        cout << ARRAY_SIZE(dArray) << endl;
       }

Solution 1

How would I find the size in bytes

This tells you how many elements are present in an array. And it gives the expected answer of 2 for dArray.

#define ARRAY_SIZE(array) (sizeof((array))/sizeof((array[0])))

If what you want it the byte count, this will do it.

#define ARRAY_SIZE(array) (sizeof(array))

Maybe because the array is only declared and not filled with anything?

That won't affect the behavior of sizeof. An array has no concept of filled or not filled.

Solution 2

array[0] contains 5 * 3 * 4 elements so sizeof(array) / sizeof(array[0]) will give you 2 when the first dimension of your array is 2

Rewrite your macro as:

#define ARRAY_SIZE(array, type) (sizeof((array))/sizeof((type)))

to get the number of elements,

or simply sizeof(array) to get the number of bytes.

Comments

[dyp]

Is there a reason for the double brackets in the first macro ((array))?

[Drew Dormann]

@DyP No functional reason that I'm aware of. It's the code presented in the question. :)

[djechlin]

Pretty sure OP is having more trouble getting 2*5*3*4 than converting from elem to bytes.

[dyp]

One could argue the name of the first macro would be clearer if it was ARRAY_LENGTH (or ARRAY_EXTENT). Also, there's the C++11 approach using std::extent, like template<class T> constexpr std::size_t get_extent(T const& a) { return std::extent<T>::value; }

[djechlin]

OIC. Nevermind, this is right - problem is array[0] was giant so OP should have been dividing by sizeof array[0][0][0][0] to get elem count.