C style strings, Pointers, arrays

Question

I'm having trouble understanding what a C-style string is. Happy early New Year

What I know: A pointer holds a memory address. Dereferencing the pointer will give you the data at that memory location.

int x = 50;
int* ptr = &x;    //pointer to an integer, holds memory address of x

cout << "&x: " << &x << endl;  //these two lines give the same output as expected
cout << "ptr: " << ptr << endl;

cout << "*ptr: " << dec << (*ptr) << endl;  //prints out decimal number 50
                                           //added dec, so the program doesnt 
                //continue to printout hexidecimal numbers like it did for the 
                 //the memory addresses above
cout << "&ptr: " << &ptr << endl;  //shows that a pointer, like any variable,
                                  //has its own memory address

Now to what I don't understand (using what's above as the source for my confusion): There are multiple ways to declare strings. I'm learning C++, but you can also use C-style strings (good to understand, although inferior to C++ strings)

C++:

string intro = "Hello world!"; 
//the compiler will automatically add a null character, \0, so you don't have to
//worry about declaring an array and putting a line into it that is bigger than 
//it can hold. 

C-style:

char version1[7] = {'H','i',' ','y','o','u','\0'};
char version2[] = "Hi you"; //using quotes, don't need null character? added for you?
char* version3 = "Hi you";

Version3 is where I'm having trouble. Here, there is a pointer to a char. I know that an array name is a pointer to the first element in an array.

cout << " &version3: " << &version3 << endl; //prints out location of 'H'
cout << " *version3: " << *version3 << endl; //prints out 'H'
cout << "  version3: " <<  version3 << endl; //prints out the whole string up to
                                             //automatically inserted \0

Before, in the section "what I knew," printing out the name of a pointer would print out the address that it held. Here, printing out the name of a pointer prints out the whole string. Do the double quotes around "Hi you" somehow tell the program: "hey I know you are a pointer, and you are initialized to the location of 'H', but because I see these double quotes, skip forward 1 byte in memory location and printout everything you see until you reach a \0" (1 byte movement because char's are 1 byte large).

How is it that printing out a pointer prints out a string? Before printing out a pointer name printed out the memory address it was initialized to.

Edit: Does cout << &version3 print out the location of 'H' or the location of the pointer, version3, which holds the memory address of 'H'?