I don't quite understand where the error is here:
int *parr[22]; // Array of int* pointers
parr[0] = ptr1;
parr[1] = ptr2;
//...
int *(*pparr)[22]; // A pointer to a int* array[22]
pparr = parr; // ERROR
the error tells me error C2440: '=' : cannot convert from 'int *[22]' to 'int *(*)[22]'
how come that the types are not equal? The name of the array should be equal to a reference to the first element of the array, something like
parr => &parr[0]
so the line seems right to me
As pparr is A pointer to a int* array[22] so you need to write
pparr = &parr;
You need to store address in the pointer and not the pointer itself.
It is same like when you have
int a=3;
int *b;
b=&a;
You are storing address of a in b, similarly you need to store address of parr in pparr
EDIT: To clarify OP's comment
You can't assign the address of the first element, but the address of the pointer that is pointing to first element.(therefore pparr = &parr;)
pparr = &parr[0] we are writing pparr=&parrAn int*[22] can decay to an int**, but you cannot assign an int** to an int*(*)[22].
int *(*pparr)[22]; //This one is an array of function-pointers returning an int pointer.
int **pptr; //Points to an array of pointer
So you can write
pptr = parr;
pparr = &parr;for the types to be compatible.