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