Address of an array different with the address of the first element?

As I know address of array a is the address of first element of this array.

void func(int a[])
{
    cout << "address in func: " << &a << endl;;
    cout << "GT: " << &a[0] << endl;
}
int main ()
{
    int a[] = {0,1,2,3};
    cout << "address in main: " << &a << endl;
    cout << "address in main a[0]: " << &a[0] << endl;

func(a);
}

Output:

address in main: 0x7ffef67d6790

address in main a0: 0x7ffef67d6790

address in func: 0x7ffef67d6778

GT: 0x7ffef67d6790

Why address of array a in func() difference with address of a[0]?

>Solution :

Why address of array a in func() difference with address of a[0]?

Because you’re calling the function func() passing a by value. This means the a inside the function func() is actually a copy of the original decayed pointer that you’re passing to func() from main(). And since they’re different variables, they have different addresses. Thus by writing cout << &a; you’re printing the address of this separate local variable named a.

If you want to print the original address, you should write inside func():

void func(int a[])
{
//---------------------------------v---------->removed the &
    cout << "address in func: " << a << endl;;
    cout << "GT: " << &a[0] << endl;
}

Demo

address in main: 0x7ffcfb0a4320
address in main a[0]: 0x7ffcfb0a4320
address in func: 0x7ffcfb0a4320
GT: 0x7ffcfb0a4320