指针数组和指向指针数组的指针

Question

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

Answer 1

int*[22]可以衰减为int** ，但不能将int**分配给int*(*)[22] 。

Answer 2

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; )

Answer 3

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;