常数和指针

Question

This is going to be a long-winded post, so I apologize in advance.. but trying to ask multiple questions with context:

So, I've been trying to learn C++, and I've been picking everything up fairly easily, but I make a habit of trying to do things alongside the tutorials in "slightly different ways", in order to ensure that I understand the functionality of the thing I'm learning.

I seem to have hit a mental roadblock regarding constants and their involvements with pointers.

In the previous section, I think I got a pretty solid grip on references, dereferences (and the difference between * when used for dereferences and pointers), and then wrote this code:

int a[5];
int * p;

p = a;

cout << "Select the value for the first item at address(" << &a[0] << ")" << nl;
cin >> *p;
p++;
cout << "Select the value for the second item at address(" << &a[1] << ")" << nl;
cin >> *p;
p++;
cout << "Select the value for the third item at address(" << &a[2] << ")" << nl;
cin >> *p;
p++;
cout << "Select the value for the fourth item at address(" << &a[3] << ")" << nl;
cin >> *p;
p++;
cout << "Select the value for the fifth item at address(" << &a[4] << ")" << nl;
cin >> *p;

for (int n=0;n<5;n++)
    cout << a[n] << "(" << &a[n] << ")" << nl;

Now, I realize that this isn't necessarily the most "optimized" solution, ever.. but it worked, I was able to "write" into the array I had store on the memory, and display the changes. It was neat, and like I said.. I think I got a pretty good understanding of how these things worked.

..and then, I hit the next section, which used the example:

const char * terry = "hello";

to teach me, and I quote: "As in the case of arrays, the compiler allows the special case that we want to initialize the content at which the pointer points with constants at the same moment the pointer is declared" (this being on the cplusplus.com tutorial ).

Anyway, to get to the point: in order to try and familiarize myself with it and the previous section, I tried writing:

const char * p = "Hello.";

for (int n=0;n<5;n++)
    cout << p[n] << "(" << &p[n] << ")" << nl;
    cout << nl;

..and this does not work. I essentially tried replicating my previous code (although without the writing aspect) in an attempt to be able to read the RAM-location of each character in this constant array, and I experimented with every possible combination of pointers, references, dereferences, etc.. but I just get a print out of the whole "Hello." array with &p.

How in the hell does one get the reference of a constant array? I think I understand the basic premise that I'm creating an array and immediately allocating it to memory, and then assigning a pointer (p) to the first block ("H")'s location.. but I can't seem to print the blocks out. I was able to get the location of the first block to print at one point (can't remember what code I used to do this) and tied it into the for loop, but even then.. it just ALWAYS printed the location of the first block, even when assigned to follow the [n] count.

I have no idea what I'm missing, here, or if I'm just trying to reach "over my head", but maybe I just don't understand the usage/point of creating an array without being able to location the memory assignment of each item within it..

This brings me to my next issue, though. How does one get the memory location of a char variable, whether constant or not? I have been having issues with this code, as well:

char const a = 100;
int * b = &a;

cout << a << nl << *b << nl;
cout << nl;

Obviously, this doesn't work.. because I can't assign b to the location of a, since they're different variables.. and obviously making a char pointer is silly, since it's just going to print messed up/null characters instead of the location (I checked this, obviously.. not just assuming).

I hope I'm just missing an option to use which may just be "more advanced" than where I'm currently at.. but if I'm not, I don't quite see the point in being able to access the block location of any type of variable/constant, if you can't access ALL of them due to restrictions of the type being used.

This problem is sorta what limits the third quoted code, too: the fact that I can't just assign a variable to hold the &p value, since it itself would have to be a char to be compatible.. which is pointless since char can't hold the eight-digit reference.

Answer 1

In

const char * p = "Hello.";

[...]   cout << &p[n]   [...]

&p[n] is a pointer to char , and cout::operator<< has an overload that takes a pointer to a char and prints a null-terminated string, that's probably why you're confused.