I have the following declaration:
char ***a; a = new char**[1]; a[0] = new char*[2]; a[0][0] = "Dynamic"; a[0][1] = "Array"; Now I need to find the number of elements in "a" as well as in "a[0]". How can I do this in C or C++?
- 1Related: stackoverflow.com/a/8320324/951890Vaughn Cato– Vaughn Cato2013-10-05 16:57:02 +00:00Commented Oct 5, 2013 at 16:57
- possible duplicate of Is this a good way to find the length of a dynamically allocated array?SJuan76– SJuan762013-10-05 16:58:12 +00:00Commented Oct 5, 2013 at 16:58
- no_of_elements = sizeof( array ) / sizeof( array[0] );Vandesh– Vandesh2013-10-05 16:58:17 +00:00Commented Oct 5, 2013 at 16:58
- 2@Vandesh No, there is no way to get size of a dynamic allocation from a pointerAnycorn– Anycorn2013-10-05 17:05:29 +00:00Commented Oct 5, 2013 at 17:05
- 1You cannot do that. This is one of the many drawbacks of dynamically allocated plain arrays.juanchopanza– juanchopanza2013-10-05 17:24:43 +00:00Commented Oct 5, 2013 at 17:24
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1 Answer
As far as I know there is no way to find number of elements in dynamically allocated array, when you pass the first element of the array as pointer to some function/method. Best practice is to avoid usage of such arrays, and when you use it, pass the number of allocated elements together with the pointer.
void doSomething(int * p, int elms) { //... } int main(){ int * arr = new int[10]; doSomething(arr, 10); }