How do I fix the size of my array? - c++

Question

everyone. I am struggling with understanding why my numbers variable keeps outputting 1. I am trying to double the size anytime the number of elements is equal to the size of the array but I end up not getting anywhere near that part of my code.
#include <iostream>
#include <string>
using namespace std;
int main()
{
int selection;
int size = 2;
// Initializing the array
int *dArray;
int v = 0;
int i, j, temp;
dArray = new int[size];
dArray[0] = 2;
dArray[1] = 3;
int numbers = sizeof(dArray) / sizeof(dArray[0]);
do {
// Printing the menu
cout << "1) Print Elements" << endl;
cout << "2) Add Element" << endl;
cout << "3) Delete Element" << endl;
cout << "4) Return Size" << endl;
cout << "5) Exit" << endl;
cout << "Enter your selection number: ";
cin >> selection;
switch (selection)
{
case 1:
// Outputting the elements
for (int i = 0; i < size-1; i++)
{
cout << dArray[i] << ", ";
}
cout << dArray[size - 1] <<endl;
cout << numbers << endl;
break;
case 2:
// Asking for another element
cout << "What number shall be put into the array? \n";
cin >> v;
if (numbers== size)
{
// If the size is too small...
int *nArray = new int[2 * size];
for (int i = 0; i < size; ++i)
{
nArray[i] = dArray[i];
}
delete[] dArray;
dArray = nArray;
// Finished creating a new array
cout << "Array size expanded to " << 2 * size << endl;
// Adding the element
dArray[size] = v;
size = 2 * size;
// Sorting the elements
for(i=0;i<numbers;i++)
{
for(j=i+1;j<numbers;j++)
{
if(dArray[i]>dArray[j])
{
temp =dArray[i];
dArray[i]=dArray[j];
dArray[j]=temp;
}
}
}
}
else
{
// Adding the element
dArray[size] = v;
size = 2 * size;
// Sorting the elements
for(i=0;i<numbers;i++)
{
for(j=i+1;j<numbers;j++)
{
if(dArray[i]>dArray[j])
{
temp =dArray[i];
dArray[i]=dArray[j];
dArray[j]=temp;
}
}
}
}
break;
}
} while (selection!= 5);
cin.get();
cin.ignore();
return 0;
}
Does anyone know why the sizeof function keeps acting this way?

Answer 1

This is not the size of your array. It's the size of a pointer, divided by the size of an int.
int numbers = sizeof(dArray) / sizeof(dArray[0]);
This is the size of the array in your program.
int numbers = size;

Answer 2

sizeof(dArray) gives you the size of dArray. dArray is defined as int *dArray, so its size is sizeof(int*). That doesn't change, regardless of what the pointer points to. That's different from an actual array; your size code would work correctly with int dArray[3];.
To get the size of the array, just use your size variable; that's the number of int objects that were allocated.
Even better, use std::vector<int>; its size() member function tells you how many elements it has.

Answer 3

int numbers = sizeof(dArray) / sizeof(dArray[0]);
Numbers is resolving to 1 because dArray is a pointer to the start of your array, not the array itself. When you call sizeof() you are getting the size of the pointer, not the array. On a 32 bit application, the pointer is 32 bits (4 bytes) and your first int is also 32 bits (4 bytes). So what this statement resolves to is
int numbers = 4 / 4;
leaving 1. If you instead declared an array of doubles then sizeof(dArray[0]) would be 8 but the size of the pointer would still be 4, so the statement would resolve to numbers = 4 / 8 = .5, but since numbers is an int it would just resolve to 0. So your code would still break, but in new exciting ways.
Interestingly enough, if you compiled this as a 64 bit application sizeof(dArray) would be 64 bits (8 bytes), so you would get numbers = 8 / 4 = 2 as you expected. Then you would double your array size and your code would break because numbers would resolve to 2 which would potentially be even more confusing, so be thankful you caught it now!
I agree with what others have said though. If you use a Vector you can just keep pushing values onto the end of it and let it worry about changing the size of the array!