什么是指针类型以及如何使用它？

Question

okay so i have come across this pointer type and i tried to use it in my code but i get a warning from the compiler saying its in incompatible pointer type

this is the type

data_type (*i)[j] with i being the variable name and j being the size of the pointer

for example if you want the pointer to be as size of 4 ints you would declare

int (*i)[4] and then you would need to assign an array of 4 ints

i = &s[4]

however when i try to assign it, i get a warning from compiler saying incompatible type

so what seems to be the problem here? and how do i use it correctly?

Answer 1

cdecl.org tells us that it is:

 int (*i)[4] 

declare i as pointer to array 4 of int

So, we can use it like this:

int arr[4];
int (*i)[4] = &arr;

Or, with heap allocated memory:

int (*j)[4] = malloc(sizeof(int[4]));

Answer 2

int (*i)[4]; declares i to be a pointer to an array of 4 int s. You said s is an array of 4 int

int s[4];  

assignment

i = &s[4];  

is wrong because &s[4] is the address of memory block just past the array and is of type int * . Generally arrays decay into pointer to its first element but in case of as an operand of unary & operator it doesn't decay and therefore &s gives the address of array s which is of type int (*)[4] .

i = &s;

Answer 3

If j is variable, then I'd recommend to use std::vector instead and take a reference to the container rather than a pointer:

std::vector<int> i(j);
auto& i_ref = i;

If j is a constant expression, I'd still avoid using C-style arrays and use std::array instead, and still take only a reference:

constexpr std::size_t j = 4;
std::array<int, j> i;
auto& i_ref = i;

Answer 4

This

int s[4];

declares s to be array of 4 int s.

Using the expression s[4] (after s had been defined) would evaluated to the 5th element of s (an int ).

As s only has 4 elements, doing so is "illegal" and provokes undefined behaviour of the program.

So to assign to int(*i)[4] the address of s do:

 i = &s;

If using C99 or later to initialise i without declaring s you might also do:

i = &(int[4]){0};