When we run this in C++

int main(){
int array[5] = {1,2,3,4,5};
cout << *(&array);
}

Why does it put address of first element of array? I thought that (&array) would give address of first element of array and *(&array) would give VALUE inside it because we are using dereferene * operator on a pointer.

But it shows address of first element of array.

Please can someone tell why it is so?

Thanks

Expected

Value at address in first array index(value of first array element)

>Solution :

There are a number of things happening here that can easily give someone a migraine, so let’s try to look at it from a much simpler approach. Let’s just say that all you have:

int p;

Then, it’s pretty simple to understand: *&p is the same thing as p. Taking an address of something, then dereferencing it, brings you right back where you started. You have accomplished absolutely nothing at all, whatsoever.

*(&array);

This is the same exact thing as: *&array, or array, or the address of the first value in the array. The End.