Usage of * in a char array

Question

int a;
char *arr[] = {"c", "c++", "java", "vba"};
char *(*ptr)[4] = &arr;
cout << ++(*ptr)[2];

I am comparing line 3 of the above code with int*p = &var.
Is it correct to say that the first * in line 3 is of the same idea with the one seen in int*p = &var?
What, then, is the purpose of the second * seen in line 3? Is it part of the pointer variable's name?

In the 4th line, how come the * outside of the parenthesis is not included?

And finally, does the ++ operator print the "a" or "ava" of (*ptr)[2]? Why?

Answer 1

Solus let's see what is happening at each and every line of code.

here in line 2 you have declared a 2-D array of strings which is nothing but an array of pointers pointing at starting of each string.

In line 3 you create another array of pointers and that is pointing to starting of your arr[].

Line 4 you are accessing the pointer that is pointing to the third element( (*ptr)[2] ) of array (which is again a pointer pointing to letter 'j' in "java")
And as soon as you increment this one(using ++ operator) , it starts pointing at 'a' of "java"

Thus the output is "ava"

Hope it is much clear now but if you are still confused then do let me know, I will try to simplify it a bit further.🙂👍