How to print size of array parameter in C++?

How to compute the size of an array CPP?

C++

// 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:

• Do Not Use sizeof For Array Parameters in C
• Why Does C Treat Array Parameters as Pointers?

How to find the size of an array in function?

We can pass a ‘reference to the array’.

C++

// 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.

C++

// 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

C++

// 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.

CPP

#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 Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above