How to get number of elements in the array of pointers?

How to get number of elements in array of pointers?

Below is my code:

struct mystruct

int a;
char ch[10];
int c;
;

mystruct *m_arr[2];


I am traversing this array in some other files. Instead of hard-coding as 2 in every file I want to get the number of elements in the array programmatically.

std::vector

std::array<mystruct,2>

extern mystruct *m_arr[ ]

m_arr

7 Answers
7

Don't use raw arrays. Use a standard container like std::vector or std::array. Both of these have a .size() member, and allow the range-based for syntax:

std::vector

std::array

.size()

for (mystruct* p : m_arr)


If you need C compatability, they both offer a data() member function which returns a pointer to the first element in the underlying array. (In your case, that will be a mystruct **)

data()

mystruct **

Edit: A raw array also supports the range-based for syntax - but only if the visible declaration includes the number of elements (so my_struct* m_arr[2]; is fine, but my_struct* m_arr would not work). It is impossible to declare a std::array without defining the size too.
Other containers (like std::vector)
don't include the size in the declaration.

my_struct* m_arr[2];

my_struct* m_arr

std::array

std::vector

The usual way of doing this is:

size_t sizeOfArray = sizeof(m_arr)/sizeof(m_arr[0]);


About size_t from wiki:

size_t

size_t is an unsigned integer type used to represent the size of any object (including arrays) in the particular implementation. The sizeof operator yields a value of the type size_t. The maximum size of size_t is provided via SIZE_MAX, a macro constant which is defined in the header (cstdint header in C++). size_t is guaranteed to be at least 16 bits wide.

_countof

size_t sizeOfArray = _countof(m_arr)

If this is a header file included by all other source code files using m_arr you can use sizeof(m_arr)/sizeof(m_arr[0]) to obtain the number of elements in the array. But this is really dangerous. If at some point the pointer m_arr enters a function as a parameter the scope of the function will not return 2 at sizeof(m_arr) but the number of bytes which the pointer takes in the memory which probably is 8.

m_arr

sizeof(m_arr)/sizeof(m_arr[0])

m_arr

sizeof(m_arr)

So if you want to stick to plain C then you have to pass the number of elements in a separate variable. But if you can use C++ there is a variety of better, safer and even faster solutions.

You can always do such memory arithmetics:

size_t arraySize = sizeof(m_arr) / sizeof(m_arr[0]);


But if you don't have a reason like in When would you use an array rather than a vector/string?:

Then you should use a std::array<mystruct*, 2> m_arr;, the size of which you can access through m_arr.size(), and the address of the underlying native array you get with m_arr.data().

std::array<mystruct*, 2> m_arr;

m_arr.size()

m_arr.data()

In C++17 you can #include <iterator> and use std::size(m_arr).

#include <iterator>

std::size(m_arr)

Yes, use std::vector, but if you must...

this works for c++11 or greater:

template <typename T, std::size_t N>
constexpr std::size_t sizeofArray(T(&)[N])
return N;



Saw this recently in a video about 7 Features of C++ You Can Adopt Today

Assuming that you have your array as the following:

char *array[3];


Add an extra item to the array, which is considered to serve as an array terminator.

char *array[3]="First element","Second element","";


The rest of it depends on your own coding style, you can check for the terminator and count the number of elements except the terminating string.

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