Question

Possible Duplicate:
How to find the sizeof(a pointer pointing to an array)
Sizeof array passed as parameter

I have this code:

class total {

public:
    void display(int []);
};

void total::display(int *arr)
{
    int len = sizeof(arr)/sizeof(int);
    cout << len;
}

int main(void)
{
    int arr[] = { 10, 0, 0, 4, 20 };
    total A;
    A.display(arr);
    return 0;
}

The output is 1 while I expected it to be the length of array , 5
However if I use the sizeof() statement in main() it displays 5.
So, why is the sizeof() not displaying correct value inside the member function?

Answer 1

The sizeof operator returns the size of the operand. In your case sizeof(arr), the type of the operand is int*. So, the result is either 4 or 8 (depending on the platform, can be also 2 or 1). There is not way to know inside the finction the length of the passed array. Even if you write

void total::display(int arr[5])
{
}

this will not change anything because arrays are converted to pointers when they are used as params of the methods. You can still pass array of any size.