I am solving problems on online judge like Leetcode and I wonder if it' possible to get size of a 2d array given int**A. Consider the function,
int help(int** A){
int rows = sizeof(A)/sizeof(A[0]);
int columns = sizeof(A[0])/sizeof(A[0][0]);
}
But I am not getting the correct values of rows and columns. Is there a way to get sizes of a 2d array if I only have int** A. Same question for char** A. I know that the question is poorly framed but I am a beginner in C. Thank you.
There's nothing in the allocated memory that indicates where it starts and ends. For 2D arrays, there's not even a guarantee the memory is contiguous.
**A does not contain any data about itself - all of the info about the array must be kept track of by the programmer.
**A isn't a pointer to the first element in an array. It's a pointer to another pointer that points to an int. There may or may not be ints following. It might also be a pointer to an array of pointers and the first element of that array points to an int. But that's all speculation. The only thing known for sure is that it's a pointer to a pointer to an int.
**A is not "the address of a pointer to a pointer of an int". **A is actually an int. *A is a pointer to an int, and its value is the address of an int. A is a pointer to a pointer to an int - so its value may be the address of a pointer to an int.int **A which declares a pointer to a pointer to an int. You are, of course, correct that the stand alone expression **A is just an int. But I think the context is clear given that **A only appears in a declaration.**A was "the address of a pointer to a pointer of an int". It has since been edited to remove that text.