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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
This code is similar, but using a dynamic array:
#include #include template 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
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 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
