Array pointers: First look at the name "array pointer" is "pointer", it can be understood as a pointer. But there is an array in front of it, that is, it points to an array.
See-"Int (*p) [Ten], we parse from an expression, because () has a high priority,so *p on behalf of P meansThe PIN, also int is modified by [10] to indicate
An array containing 10 int, so p represents the first address of an array with 10 elements. Maybe you should define an array pointer like this: Int (*) [ten] P
because according to our method, the first is the type, followed by the variable, so the previously defined type is an array pointer, It is followed by the name of the pointer. But the compiler doesn't think so, because it's too ugly to write. So it was written. Int (*p) [Ten] array of pointers:Pointer array, so the name of the pointer array is an "array", so, but the contents of the array is a pointer, so we int *p[10] This can be understood here, because [] the priority is higher than * , so first the p and [] are combined into an array, so the array name
is P, and the following int * Modifies the contents inside, resulting in an array containing 10 int*.
Pointer
to pointersThe pointer is a pointer to what is stored inside the pointer or a pointer address.
array pointer Pointer array pointer pointer to the relationship of the shape participation argument
The " array name is rewritten as a pointer parameter" rule is not recursively defined. Arrays of arrays are rewritten as "pointers to arrays" instead of " pointer pointers":
Parameters that are matched by an argument
Array of array char c[8][10]; CHAR (*) [10]; Array pointers
Pointer array char *c[10]; Char **c; Pointer to pointers
Array pointer (row pointer) char (*C) [10]; char (*C) [10]; does not change
Pointer to the pointer char **c; Char **c; does not change
Source: >
From for notes (Wiz)
Pointer to array pointer, pointer array, and pointer
