Pointer to an array get size C++ - c++

Question

int * a;
a = new int[10];
cout << sizeof(a)/sizeof(int);
if i would use a normal array the answer would be 10,
alas, the lucky number printed was 1, because sizeof(int) is 4 and iszeof(*int) is 4 too. How do i owercome this? In my case keeping size in memory is a complicated option. How do i get size using code?
My best guess would be to iterate through an array and search for it's end, and the end is 0, right? Any suggestions?
--edit
well, what i fear about vectors is that it will reallocate while pushing back, well you got the point, i can jus allocate the memory. Hoever i cant change the stucture, the whole code is releevant. Thanks for the answers, i see there's no way around, so ill just look for a way to store the size in memory.
what i asked whas not what kind of structure to use.

Answer 1

Simple.
Use std::vector<int> Or std::array<int, N> (where N is a compile-time constant).
If you know the size of your array at compile time, and it doens't need to grow at runtime, then use std::array. Else use std::vector.
These are called sequence-container classes which define a member function called size() which returns the number of elements in the container. You can use that whenever you need to know the size. :-)
Read the documentation:
std::array with example
std::vector with example
When you use std::vector, you should consider using reserve() if you've some vague idea of the number of elements the container is going to hold. That will give you performance benefit.
If you worry about performance of std::vector vs raw-arrays, then read the accepted answer here:
Is std::vector so much slower than plain arrays?
It explains why the code in the question is slow, which has nothing to do with std::vector itself, rather its incorrect usage.
If you cannot use either of them, and are forced to use int*, then I would suggest these two alternatives. Choose whatever suits your need.
struct array
{
int *elements; //elements
size_t size; //number of elements
};
That is self-explanatory.
The second one is this: allocate memory for one more element and store the size in the first element as:
int N = howManyElements();
int *array = int new[N+1]; //allocate memory for size storage also!
array[0] = N; //store N in the first element!
//your code : iterate i=1 to i<=N
//must delete it once done
delete []array;

Answer 2

sizeof(a) is going to be the size of the pointer, not the size of the allocated array.
There is no way to get the size of the array after you've allocated it. The sizeof operator has to be able to be evaluated at compile time.
How would the compiler know how big the array was in this function?
void foo(int size)
{
int * a;
a = new int[size];
cout << sizeof(a)/sizeof(int);
delete[] a;
}
It couldn't. So it's not possible for the sizeof operator to return the size of an allocated array. And, in fact, there is no reliable way to get the size of an array you've allocated with new. Let me repeat this there is no reliable way to get the size of an array you've allocated with new. You have to store the size someplace.
Luckily, this problem has already been solved for you, and it's guaranteed to be there in any implementation of C++. If you want a nice array that stores the size along with the array, use ::std::vector. Particularly if you're using new to allocate your array.
#include <vector>
void foo(int size)
{
::std::vector<int> a(size);
cout << a.size();
}
There you go. Notice how you no longer have to remember to delete it. As a further note, using ::std::vector in this way has no performance penalty over using new in the way you were using it.

Answer 3

If you are unable to use std::vector and std::array as you have stated, than your only remaning option is to keep track of the size of the array yourself.
I still suspect that your reasons for avoiding std::vector are misguided. Even for performance monitoring software, intelligent uses of vector are reasonable. If you are concerned about resizing you can preallocate the vector to be reasonably large.