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; } |
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’.
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; } |
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; } |
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; } |
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
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