Question

here is some code

class DengkleTryingToSleep{
public:
int minDucks(int ducks[]);
int temp(int ducks[]){
int size=sizeof(ducks);
cout<<"sizeof="<<size<<"\n";
}
};


int main (int argc, const char * argv[])
{
DengkleTryingToSleep dt;
int arr[]={9,3,6,4};

cout<<"sizeof "<<sizeof(arr);
cout<<"\nsizeof from function "<<dt.temp(arr);

return 0; 
}

and output of this is

sizeof 16
sizeof from function sizeof=8

and i have no idea how this is working because it returns 16 (as expected when called inside main)
and returns 8 when called from the function

Answer 1

Actually this function:

int temp(int ducks[])

is exactly equivalent this function:

int temp(int *ducks)

There is NO DIFFERENCE at all. No difference. So no matter what you pass, whether an array or a pointer, it will become a pointer inside the function.

That means, when you write sizeof(ducks) in your function, it is exactly equivalent to sizeof(int*), which returns 8 on your machine (I guess, your machine has 64-bit OS where the size of pointer is 8 bytes).

If you want to pass an array, and don’t it decay into pointer type, then do this:

template<size_t N>
int temp(int (&ducks)[N])
{
    int size=sizeof(ducks);
    cout<<"sizeof="<<size<<"\n";
}

Now it will print 16. Note that inside the function N represents the count of items in the array. So in your case, it would be 4, as there are 4 elements in the array. It means, if you need the length of the array, you don’t need to calculate it as sizeof(bucks)/sizeof(int), as you already know the length of the array which is N.

Also note that there is a limitation in this approach: now you cannot pass dynamically allocated array:

int *a = new int[10];
dt.temp(a); //compilation error

//but you can pass any statically declared array
int b[100], c[200];
dt.temp(b); //ok - N becomes 100
dt.temp(c); //ok - N becomes 200

---

But in C++, you’ve a better option here: use std::vector<int>.

int temp(std::vector<int> & ducks)
{
     std::cout << ducks.size() << std::endl;
}

//call it as
std::vector<int> v = {1,2,3,4,5,6}; //C++11 only, or else : use .push_back()
dt.temp(v);