C++ pointer to an array of ints

Question

int main(){
    int a[4] = { 1,2,3,4 };
    int(*b)[4] = &a; //with a doesn't work
    cout << a << &a  << endl; //the same address is displayed
}

So, int(*b)[4] is a pointer to an array of ints. I tried to initialize it with &aand a both. It works only with the first.
Aren't they both addresses of the first element of the array?

Pointers have a type; (i.e. they're an address and a type). This doesn't work for the same reason that int x; double *p = &x; doesn't work. — M.M
– M.M, Commented Jul 20, 2017 at 2:54
@P. Danielski: How would you differentiate between "address of the first element of the array" and "address of the entire array" if both begin at the same point in memory? A similar example: struct S { int x; } s;. Then &s and &s.x are the same numerical address. So, what is &s then: adress of the entire struct object or address of its first field? — AnT stands with Russia
– AnT stands with Russia, Commented Jul 20, 2017 at 2:57

songyuanyao · Accepted Answer · 2017-07-20 02:51:29Z

7

Conceptually they're not the same thing. Even they point to the same address, they're incompatible pointer types, thus their usages are different either.

&a is taking the address of an array with type int [4]), then it means a pointer to array (i.e. int (*)[4]).

a causes array-to-pointer decay here, then it means a pointer to int (i.e. int*).

edited Jul 20, 2017 at 2:51

answered Jul 20, 2017 at 2:17

songyuanyao

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Shiu Chun Ming · Accepted Answer · 2017-07-20 06:51:01Z

0

Take a look at this piece of code

int a[4] = { 1, 2, 3, 4 } // array 
int(*b) = a // pointer to the first element of the array

You should know that int(*b) is equal to int(*b)[0] and a[0]. Therefore, It's a pointer pointing to an integer(int*), not a pointer pointing to an array of integer.

That how type issue arises in your case.

Noted that C is a strong type language. Look at your assignment statement.

int(*b)[4] = &a;

It takes a parameter of int(*ptr)[4]. It means you have to strictly pass the argument of that type, which is a int *.

And you are trying to pass a pointer to array of 4 int to int * . Therefore, they're not compatible in the assignment, even their address are the same.

edited Jul 20, 2017 at 6:51

answered Jul 20, 2017 at 6:41

Shiu Chun Ming

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