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Initializing pointer to array c++

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.

Answer*

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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Initializing pointer to array c++

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