How to get size of dynamic array in C++
c++, dynamic-arrays
Solution
You can't. The size of an array allocated with `new[]` is not stored in any way in which it can be accessed. Note that the return type of `new []` is not an array - it is a pointer (pointing to the array's first element). So if you need to know a dynamic array's length, you have to store it separately.
Of course, the proper way of doing this is avoiding `new[]` and using a `std::vector` instead, which stores the length for you and is exception-safe to boot.
Here is what your code would look like using `std::vector` instead of `new[]`:
size_t n; // Size needed for array - size_t is the proper type for that
cin >> n; // Read in the size
std::vector<int> a(n, 0); // Create vector of n elements initialised to 0
. . . // Use a as a normal array
// Its size can be obtained by a.size()
// If you need access to the underlying array (for C APIs, for example), use a.data()
// Note: no need to deallocate anything manually here
Problem
Code for dynamic array by entering size and storing it into "n" variable, but I want to get the array length from a template method and not using "n". ``` int* a = NULL; // Pointer to int, initialize to nothing. int n; // Size needed for array cin >> n; // Read in the size a = new int[n]; // Allocate n ints and save ptr in a. for (int i=0; i<n; i++) { a[i] = 0; // Initialize all elements to zero. } . . . // Use a as a normal array delete [] a; // When done, free memory pointed to by a. a = NULL; // Clear a to prevent using invalid memory reference. ``` This code is similar, but using a dynamic array: ``` #include <cstddef> #include <iostream> template< typename T, std::size_t N > inline std::size_t size( T(&)[N] ) { return N ; } int main() { int a[] = { 0, 1, 2, 3, 4, 5, 6 }; const void* b[] = { a, a+1, a+2, a+3 }; std::cout << size(a) << '\t' << size(b) << '\n' ; } ```