char array & pointers

[S6k9GNh0]

I tried to assign a pointer with arrays since I was having trouble understanding the concept:

#include <iostream>

using namespace std;

int main()
{
    char aPointer[] = "add";
    char* pCarrier[sizeof(aPointer)];
    pCarrier[sizeof(aPointer)] = aPointer;
    cout << *pCarrier;
    return 0;
}

Without using the ampersand it assigns the pointer a reference but when you try and get the value from the reference using the cout function, it gives me what looks this: Uï∞âSVWUⁿï]♀ï≈@♦♠

If I use:

#include <iostream>

using namespace std;

int main()
{
    char aPointer[] = "add";
    char* pCarrier[sizeof(aPointer)];
    pCarrier = &aPointer;
    cout << *pCarrier;
    return 0;
}

It gives me an bad type compatibility error.


I was wondering what I'm doing wrong, and how to fix it if possible. I've tried tons of combinations for experimenting but I've gotten to a point to where I just want to know what the right thing to do is. This would add to my experience and hopefully help me with the concept. Also, I'm a little sketchy with terms and I don't mind to be corrected in a nice way. :) Thank you much in advance.

[firedraco]

It's because you made pcarrier an array of arrays...(a pointer is basically the beginning of an array).

Is this what you are trying to do?

char aPointer[] = "add";
char* pCarrier = &aPointer;
cout << *pCarrier;
return 0;

[jsmith]

I'm not sure what you are trying to do, but you are definitely confusing pointers with arrays with pointers to arrays.

Let's look at what your code does.

char aPointer[] = "add";

Declares an array of 4 characters: 'a', 'd', 'd', '\0' (NULL)

char* pCarrier[sizeof(aPointer)];

Declares an array of sizeof(aPointer) pointers to characters.
sizeof(aPointer) is 4, so you create an array of 4 pointers.
This array occupies 16 bytes of memory on a 32-bit machine.

pCarrier[sizeof(aPointer)] = aPointer;

aPointer referenced by itself is the address of the first array
in memory. sizeof(aPointer) is 4. Arrays in C are indexed zero-based,
so the valid indices in an array of 4 elements are 0, 1, 2, and 3.
You assign index 4 the address of the aPointer array. This is a
memory stomp. Since you don't explicitly assign any of the other
elements of pCarrier, they are all uninitialized (random garbage).

cout << *pCarrier;

*pCarrier is the same as pCarrier[0]. Each element of pCarrier is
a char*, so this prints out the string pointed to by pCarrier[0]
which we established is random garbage.

[S6k9GNh0]

Well figured it out. I am noob it seems. The point of this is to make a pointer that pointed to an array of char and have the ability to return the array of char. If you do:

char aPointer[] = "add";
char* pCarrier = &aPointer;
cout << *pCarrier;
return 0;

It only gives you the first letter or 'a' or the reference of aPointer[0]. So you would have to make the array of pointers and assign each element the reference to that specific char since each one has its own address in memory.

#include <iostream>

using namespace std;

int main()
{
    int i;
    char aPointer[] = "I love earth and all its creatures.";
    char *pCarrier[sizeof(aPointer)];
    for (i = 0; i < sizeof(aPointer); i++)
        pCarrier[i] = &aPointer[i];
    cout << *pCarrier;
    return 0;
}

//returns the array of char in aPointer. 

Thank you guys for helping me. This really did help. :).
If any of you see a better way to do this, I wouldn't mind the advice. I'm new and I still don't have the gist of things around C++.

[jsmith]

In your example above,

char* pCarrier = &aPointer;
cout << *pCarrier;

pCarrier is a pointer to a character. It can also be interpreted as a pointer to an array of characters, or a C-style string.

The expression *pCarrier has type char -- that is, a single character -- and hence you only see one character printed. If you wanted to see the whole string printed, all you need to do is

cout << pCarrier;