Advanced Pointer convoluted syntax - c++

Question

I was fidgeting with the pointers thingy.graduated from pointers -> array of pointers -> function pointers -> pointer to pointers..
Here's what i am stuck at...mostly the convoluted syntax.
Lets say i have an array of integers.
int arr[4] = {1,2,3..};
also i have array of pointers
int* ptr[4];
ptr[0] = arr;
here ptr[0] will point to 1
and ptr[1] will point to some other location
This works perfectly !!
Now considering above background i tried this.
char* crr[4] ={"C","C++","C#","F#"};
char** btr[4];
btr[0] = crr;
which works as pointer in oth element of btr is pointing to another pointer element in crr.
Then i tried this.
char* crr[4] ={"C","C++","Java","VBA"};
char** btr[4]= &crr; // Exception: cannot convert char* [4] * to char**[4]
but when i do this it works :O
char* crr[4] ={"C","C++","Java","VBA"};
char* (*btr)[4]= &crr;
i have not understood the last two scenarios. The use of brackets on RHS Pls explain.

Answer 1

char** btr[4]= &crr; // Exception: cannot convert char* [4] * to char**[4]
That's trying to initialise an array (of pointers to pointers) from a pointer, which you can't do. If you wanted to initialise the first element to point to crr (as in your first example), then you'd do
char** btr[4]= {crr};
The last example is declaring a pointer to an array (of pointers), and initialising it from a compatible pointer.
Note that your original array should be of const char *, not char *, since it contains pointers to (constant) string literals.

Answer 2

Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
When you write
btr[0] = crr;
the expression crr has type "4-element array of char *"; since it is not the operand of a sizeof or & operator, it is converted to type "pointer to char *", or char **. This matches the type of btr[0].
However, when you write
char** btr[4]= &crr;
the expression crr is the operand of the & operator; thus, the type of the expression &crr is "pointer to 4-element array of char *", or char *(*)[4].
The parentheses are necessary because the postfix [] operator has higher precedence than the unary * operator; the expression *a[i] will always be parsed as *(a[i]). Thus,
T *a[N]; // a is an N-element array of pointer to T
T (*a)[N]; // a is a pointer to an N-element array of T
T *(*a)[N]; // a is a pointer to an N-element array of pointer to T
The same is true for pointers and functions:
T *f(); // f is a function returning a pointer to T
T (*f)(); // f is a pointer to a function returning T
T *(*f)(); // f is a pointer to a function returning pointer to T

Answer 3

Let's look at your definitions:
Array of size 4 of char*
char* crr[4] ={"C","C++","Java","VBA"};
Array of size 4 of char**
char** btr[4]
Pointer to an array of size 4 of char*
char* (*btr)[4]
Now you are trying to assign Array of size 4 of char* to Array of size 4 of char** and they are of different type, so you, obviously, have a error. Still there is another rule(which is dominated): you can't initialize array without curly braces, so array should be initialized with new data not to be a pointer to other array.
And you have absoulutely legal pointer type Pointer to an array of size 4 of char* to which your array address could be assigned