How to print size of array parameter in C++?
How to compute the size of an array CPP?
// A C++ program to show that it is wrong to
// compute size of an array parameter in a function
#include <iostream>
using namespace std;
void findSize(int arr[])
{
cout << sizeof(arr) << endl;
}
int main()
{
int a[10];
cout << sizeof(a) << " ";
findSize(a);
return 0;
}
Output
40 8
Time Complexity: O(1)
Auxiliary Space: O(n) where n is the size of the array.
The above output is for a machine where the size of an integer is 4 bytes and the size of a pointer is 8 bytes.
The cout statement inside main() prints 40, and cout in findSize() prints 8. The reason for different outputs is that the arrays always pass pointers in functions. Therefore, findSize(int arr[]) and findSize(int *arr) mean exact same thing. Therefore the cout statement inside findSize() prints the size of a pointer.
For details, refer to the following articles:
How to find the size of an array in function?
We can pass a 'reference to the array'.
// A C++ program to show that we can use reference to
// find size of array
#include <iostream>
using namespace std;
void findSize(int (&arr)[10])
{
cout << sizeof(arr) << endl;
}
int main()
{
int a[10];
cout << sizeof(a) << " ";
findSize(a);
return 0;
}
Output
40 40
Time Complexity: O(1)
Space Complexity: O(n) where n is the size of the array.
The above program isn't appealing as we have used the hardcoded size of the array parameter.
We can do it better using templates in C++
We can use templates to define the function instead of using the hardcoded size.
// A C++ program to show that we use template and
// reference to find size of integer array parameter
#include <iostream>
using namespace std;
template <size_t n>
void findSize(int (&arr)[n])
{
cout << sizeof(int) * n << endl;
}
int main()
{
int a[10];
cout << sizeof(a) << " ";
findSize(a);
return 0;
}
Output
40 40
Time Complexity: O(1)
Space Complexity: O(n) where n is the size of the array.
We can make a generic function as well
// A C++ program to show that we use template and
// reference to find size of any type array parameter
#include <iostream>
using namespace std;
template <typename T, size_t n>
void findSize(T (&arr)[n])
{
cout << sizeof(T) * n << endl;
}
int main()
{
int a[10];
cout << sizeof(a) << " ";
findSize(a);
float f[20];
cout << sizeof(f) << " ";
findSize(f);
return 0;
}
Output
40 40 80 80
Time Complexity: O(1)
Space Complexity: O(n) where n is the size of array.
Now the next step is to print the size of a dynamically allocated array.
It's your task man! I'm giving you a hint.
#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{
int *arr = (int*)malloc(sizeof(int) * 20);
return 0;
}
This article is contributed by Swarupananda Dhua
