Help with Pointers and functions!

[reidt]

Hello guys, totally new to programming so pardon me if this is a really dumb topic to ask! I understand basic pointer theories, but I can't seem to wrap my head around pointer to an array of character pointers.
Like:

typedef char * word;
word *w [5];

I know that in essence is doing:

char **w [5];

but that's about all I understand.

I found an example online that can help with making my questions clearer.

#include <iostream>
using namespace std;
void func(char **& ref_to_ptr); /* Function declaration */
int main(void)
{
    /* Declare the '2D Array' */
    char ** ptr = new char * [5];
    ptr[3] = new char[20];

    /* Put some data in the array */
    ptr[3] = "k";

    /* Print the first value on the screen */
    cout << "First value: " << ptr[3] << endl;

    /* Pass the array by reference to the function 'func()' */
    func(ptr);

    /* Again we print on the screen what's in the '2D Array' */
    cout << "Second value: " << ptr[3] << endl;

    /* Wait for the user to press ENTER */
    cin.get();

    /* Cleanup */
    delete[] ptr;

    /* Tell the Operating System that everything went well */
    return 0;
}
void func(char **& ref_to_ptr)
{
    /* This function demonstrates how to change the value of it's argument(s) */
    ref_to_ptr[3] = "tux4life";
}

My questions:
1) What actually happens in this line?
char ** ptr = new char * [5];
Is it declaring that ptr is a pointer to a char pointer and thereafter allocating an array of 5 char pointers? So essentially, ptr is pointing to the first element of the array of 5 char pointers? Not sure if I'm making any sense?! o.O

2) Also, if the above is true, then am I right to say that I can treat ptr[1] like a character array by itself? So if I
cout << ptr [1];
I would get the string that is pointed to by ptr [1]?
Then the next question would be, how do I access the individual character in that string pointed to by ptr [1]?

3) So if i
cout << *(ptr + 1);
Am I displaying ptr[1] also, since ptr is pointing at ptr[0]?

4) Why is there a need for a * after the new char? Why can't I just do
char ** ptr = new char [5];

5) How do we pass pointers and pointers to pointers and pointers to arrays into a function? For pointers to arrays, isn't array passed by reference by default? Why is the example code using the ampersand (&) to pass the 2D array?

6)

char **w [5] //declaring an array of 5 pointers to char pointers
char **w = new char * [5] // declaring a pointer to an array of 5 character pointers 

Is that correct?

[reidt]

Oh yea, and also..

7)

int **ptr
cout << **ptr << endl;  // 2 astrerix to get the value but for char only one?

char **chptr
cout << *chptr << endl; // why only one asterix is needed? 

[MiiNiPaa]

1) Yes
2) Yes, at line 8 you allocating memory for one such string: ptr[3] = new char[20];
2a)To acces individual characters you can use: ptr[3][0] //← first character of fourth string
3) Yes
4) You get type mismatch char* and char
5) Array are not passed by reference by default. You see, when you typechar arr[5] arr is a pointer, so when you call function foo(arr) it passes pointer arr by value. You can pass array(pointer) by reference to change where it points, for example to reallocate array. In your code There is no need in passing by reference at all.
6) basically yes. If it helps you: In functions void foo(char* bar) and void foo(char bar[]) are the same thing.
7) Because there is version of operator<< for char*, which outputs c-like string. If you use two dereference operators(asterisks) it will output only first character

[reidt]

Ahh! I see! Thank you so much MiiNiPaa for helping me understand better!

I got another question though,
- why do you say that in the above code there is no need to pass by reference?
Please see if my understanding is accurate.

Is it because that ptr is already pointing to the actual location of the character array?
So if I wanted to change the string on the 3rd character array (or 3rd char pointer) I can do

 
void function (char **ptr2)
{
ptr2 = ptr2 + 2;   // in this case changing where ptr2 points is just changing a copy 
// and ptr is not affected at all?
*ptr2 = "a new string";
// or I can also use the strcpy function?
strcpy (*ptr2, "a newer string");
}

Please correct me if I'm wrong!

Also while on the strcpy topic, I read somewhere that with char pointers, when I compile there won't be any error, but as I run the program there will be a runtime error of some sort. Does that apply here if I use the strcpy function?

Thank you so much for your help!

[MiiNiPaa]

- why do you say that in the above code there is no need to pass by reference?

Because code does't changes ref_to_ptr

So if I wanted to change the string ...

Yes. ptr2 is a copy so you can change it without affecting rest of program. But you shouldn't assign string literals to c-strings like that. Use strcpy / strncpy functions.

Also while on the strcpy topic

There could be a runtime error if there is buffer overflow and strcpy is not safe too. You can use strncpy function:
http://cplusplus.com/reference/cstring/strncpy/