指向char数组的指针，指针的值不是地址吗？

Question

Here is a simple code I wrote.The pointer p 's value is the address of array a as we know.
However , why does the pointer s not store the address of c1?
How does it work!

int main(int argc, const char * argv[])
{
    int a[4] = {4,3,2,1};
    int*p = a;
    cout<<&a<<endl;//output 0x7fff5fbff8a0
    cout<<p<<endl; //oupput 0x7fff5fbff8a0

    char c1[4] = "abc";
    char *s = c1;
    cout<<&c1<<endl;//output 0x7fff5fbff894
    cout<<s<<endl; //output abc
    return 0;
}

Answer 1

why does the pointer s not store the address of c1

It does.

What you're seeing is that fact that std::ostream::operator<< has an overload for char* , treating it as a string rather than a pointer. If you use

printf("%p\n", s);

you'll see it works as expected.

Answer 2

It called operator overloading:

//char* goes here:
std::ostream& operator<<(std::ostream &s, const char* p)
{ 
  //print the string
}

//int* goes here:
std::ostream& operator<<(std::ostream &s, const int* p)
{ 
  //print the address
}

if you cast the pointer to int you'll see the address.