Initializing pointer to array c++

Question

I am new to C++. Please see the following code:

//sample1
int arr[2]  = { 11, 22 };
int(*t)[2];
t = &arr;
cout << *t[0] << "\n";
cout << *t[1] << "\n";


//sample2
int arr2[2];
arr2[0] = { 33 };
arr2[1] = { 44 };
int(*t2)[2];
t2 = &arr2;
cout << *t2[0] << "\n";
cout << *t2[1] << "\n";


//sample3
int arr3[2];
arr3[0] = { 55 };
arr3[2] = { 66 };
int(*t3)[2];
t3 = &arr3;
cout << *t3[0] << "\n";
cout << *t3[1] << "\n";

// output

Can anyone tell me how to initializing a pointer to array？Sample3 is out of my understanding.

There's nothing wrong with the initialization of arr. You should read up on the precedence of the * and [] operators. (hint: it should be (*arr)[0] etc.) — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Oct 18, 2014 at 14:45

Marco A. · Accepted Answer · 2014-10-18 15:02:21Z

There are two issues with your code:

1) [] has more precedence than * so you need parenthesis in all those cases (cfr. operator precedence). If you don't use them, you're going to do pointer-arithmetic on arrays of 2 integers each (thus immediately getting out-of-range), e.g.

int(*t)[2]; // A pointer to an array of 2 integers
cout << t[0]; // The address of the array

[first_integer][second_integer] ... garbage memory ...
^

cout << t[1]; // The address of the array + sizeof(int[2])

[first_integer][second_integer] ... garbage memory ...
                               ^
cout << *t[0]; // Dereference at the address of the array
cout << *t[1]; // Dereference past the end of the array

//  ---- correct -----

cout << (*t)[0]; // Dereference the pointer to the array and get the element there

[first_integer][second_integer] ... garbage memory ...
^

cout << (*t)[1]; // Dereference the pointer to the array and get the second element there

[first_integer][second_integer] ... garbage memory ...
               ^

2) You have an out-of-range access at line

arr3[2] = { 66 };

This is how you should proceed:

//sample1
int arr[2] = { 11, 22 };
int(*t)[2];
t = &arr;
cout << (*t)[0] << "\n";
cout << (*t)[1] << "\n";


//sample2
int arr2[2];
arr2[0] = { 33 };
arr2[1] = { 44 };
int(*t2)[2];
t2 = &arr2;
cout << (*t2)[0] << "\n";
cout << (*t2)[1] << "\n";


//sample3
int arr3[2];
arr3[0] = { 55 };
arr3[1] = { 66 };
int(*t3)[2];
t3 = &arr3;
cout << (*t3)[0] << "\n";
cout << (*t3)[1] << "\n";

The initialization is just fine.

Adrian May · Accepted Answer · 2014-10-18 14:54:10Z

1

An array is almost the same thing as a pointer:

int arr[2]  = { 11, 22 };
int * t;
t = arr;
cout << t[0] << "\n";
cout << t[1] << "\n";

answered Oct 18, 2014 at 14:54

Adrian May

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Comments

Vlad from Moscow · Accepted Answer · 2014-10-18 14:55:57Z

In this code snippet

//sample3
int arr3[2];
arr3[0] = { 55 };
arr3[2] = { 66 };
int(*t3)[2];
t3 = &arr3;
cout << *t3[0] << "\n";
cout << *t3[1] << "\n";

there is defined an array with two elements. The valid range of indices for elements of this array is 0, 1 In this statement

arr3[2] = { 66 };

there is an attempt to write data beyond the array because index 2 is invalid. So you have

arr3: | 55 | uninitilaized | 66 |
index:   0        1          beyond the array

After these statements

int(*t3)[2];
t3 = &arr3;

pointer t3 points to array arr3,

Expression

t3[0]

gives array arr3

Expression

*t3[0]

gives the first element of the array. Thus statement

cout << *t3[0] << "\n";

outputs 55.

Expression

t3[1]

points to memory after (beyond) array arr3. You wrote at this address value 66. So this value is outputed by statement

cout << *t3[1] << "\n";

Of course this code snippet is invalid because its overwrite memory that was not reserved for objects of the code.

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