How to print size of array parameter in C++? - GeeksforGeeks
How to print size of array parameter in C++?
- Difficulty Level : Medium
- Last Updated : 17 Aug, 2021
How to compute size of an array parameter in a function?
Consider below C++ program:
- 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
The above output is for a machine where size of integer is 4 bytes and size of a pointer is 8 bytes.
The cout statement inside main prints 40, and cout in findSize prints 8. The reason is, arrays are always passed pointers in functions, i.e., findSize(int arr[]) and findSize(int *arr) mean exactly same thing. Therefore the cout statement inside findSize() prints size of a pointer. See this and this for details.
How to find size of array in function?
We can pass a ‘reference to the array’.
- CPP
// 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
The above program doesn’t look good as we have hardcoded size of array parameter. We can do it better using templates in C++.
- CPP
// 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
We can make a generic function as well:
- CPP
// 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 4080 80
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;
}
