For instance:
int* pArray;
pArray = new array[];
instead of:
int* pArray;
pArray = new array[someNumber];
Since pointers are able to dynamically change the size of an array at run time, and the name of the pointer points to the first element of an array, shouldn't the default size be [1]? Does anyone know what's happening behind the scene?
Since pointers are able to dynamically change the size of an array at run time
This is not true. They can't change the size unless you allocate a new array with the new size.
If you want to have an array-like object that dynamically changes the size you should use the std::vector.
#include<vector>
#include<iostream>
...
std::vector<int> array;
array.push_back(1);
array.push_back(2);
array.push_back(3);
array.push_back(4);
std::cout << array.size() << std::endl; // should be 4
When you create an array with new, you are allocating a specific amount of memory for that array. You need to tell it how many items are to be stored so it can allocate enough memory.
When you "resize" the array, you are creating a new array (one with even more memory) and copying the items over before deleting the old array (or else you have a memory leak).
Quite simply, C++ arrays have no facility to change their size automatically. Therefore, when allocating an array you must specify it size.
Pointers cannot change an array. They can be made to point to different arrays at runtime, though.
However, I suggest you stay away from anything involving new until you have learned more about the language. For arrays changing their size dynamically use std::vector.
Pointers point to dynamically allocated memory. The memory is on the heap rather than the stack. It is dynamic because you can call new and delete on it, adding to it and removing from it at run time (in simple terms). The pointer has nothing to do with that - a pointer can point to anything and in this case, it just happens to point to the beginning of your dynamic memory. The resizing and management of that memory is completely your responsibility (or the responsibility of the container you may use, e.g. std::vector manages dynamic memory and acts as a dynamic array).
They cannot change the size dynamically. You can get the pointer to point to a new allocation of memory from the heap.
Behind the scenes there is memory allocated, a little chunk of silicium somewhere in your machine is now dedicated to the array you just newed.
When you want to "resize" your array, it is only possible to do so in place if the chunk of silicium has some free space around it. Most of the times, it is instead necessary to reserve another, bigger, chunk and copy the data that were in the first... and obviously relinquish the first (otherwise you have a memory leak).
This is done automatically by STL containers (like std::vector or std::deque), but manually when you yourself call new. Therefore, the best solution to avoid leaks is to use the Standard Library instead of trying to emulate it yourself.
int *pArray = new int; can be considered an array of size 1 and it kinda does what you want "by default".
But what if I need an array of 10 elements? Pointers do not have any magical abilites, they just point to memory, therefore:
pArray[5] = 10; will just yield a run-time error (if you are lucky).
Therefore there is a possibility to allocate an array of needed size by calling new type[size].
