Why is the size of array not same in both the cases?

Question

#include<iostream>
using namespace std;
void f(int arr[])
{
    int a=sizeof(arr);
    cout<<a;
}
int main()
{
    int n;
    cin>>n;
    int arr[n];
    int a=sizeof(arr);
    cout<<a<<"\n";
    f(arr); 
}

Output:

24
8

Why is my output not same in both the cases even when i am printing the size of same array?

Aniruddha Sarkar · Answer 1 · 2016-01-10T08:10:29.807

0

In the first call it shows the size of array. But in the second case it shows the size of the pointer passed.

When arrays are passed through functions they decay into pointers. So after passing, the:

sizeof(array)

Shows the size of the pointer.

edited Jan 10 '16 at 08:10

answered Jan 10 '16 at 08:03

Aniruddha Sarkar

1,903
2
14
15

arr is pointer to first element of the array.So when you apply it to sizeof() inside main, it should return the size of pointer not array – Abhishek Singla Jan 10 '16 at 11:01
No, in main, the variable arr is an array but in function f, the variable arr is a pointer. – Aniruddha Sarkar Jan 10 '16 at 15:59
In case of main arr is of a different datatype (an array) while in f arr is of another datatype(a pointer) altogether, here. – Aniruddha Sarkar Jan 10 '16 at 16:01

Why is the size of array not same in both the cases?

1 Answers1