Question

What is the output of the following code? int *p; int *q; p = new int; q = p; *p = 46; *q = 39; cout << *p << " " << *q << endl;

Short answer

The output is `39 39`. Both pointers point to the same memory location, which contains `39`.

Step-by-step solution

Understand Pointer Initialization

In the code, we declare two integer pointers, `int *p;` and `int *q;`. These pointers are initially pointing to undefined memory locations. They need to be allocated specific memory locations through dynamic memory allocation or by pointing to existing variables.

Dynamic Memory Allocation to `p`

We then allocate memory dynamically to `p` using the line `p = new int;`. This allocation provides `p` with a new memory location where we can store an `int` value. At this point, `p` has its own address for an integer value, but the integer itself isn't assigned yet.

Pointer Assignment `q = p;`

Next, we assign `p` to `q` with the statement `q = p;`. This means that `q` is now pointing to the same memory location as `p`. Any changes made to the value at this memory address will be reflected through both `p` and `q`.

First Value Assignment `*p = 46;`

Here, we assign the integer value `46` to the memory location pointed to by `p` using `*p = 46;`. Since `q` points to the same location as `p`, this value is accessible through both `*p` and `*q`, so both equal `46` at this stage.

Second Value Assignment `*q = 39;`

In this step, the value at the shared memory location is assigned `39` using `*q = 39;`. This overwrites the previous value `46`. Now, both `*p` and `*q` reflect the value `39`.

Output the Values

Finally, we print the values `*p` and `*q` with `cout << *p << " " << *q << endl;`. Since the value at the memory address pointed by both `p` and `q` was changed to `39`, the output will be `39 39`.

Key concepts

Dynamic Memory Allocation

Dynamic memory allocation in C++ allows you to request memory from the system at runtime. The key advantage here is that you don't need to specify the size of your data structures at compile time.
When you use `new` to allocate memory, the system searches for a suitable block of memory in the heap, allocates it, and returns its address. This is why in our code, when we use `p = new int;`, a memory space is reserved, and `p` is set with an address to this space.

• This dynamically allocated memory persists until it is manually freed using `delete`.
• Without freeing the memory, you risk memory leaks since the reserved space remains allocated and unavailable for other processes.

In essence, dynamic memory allocation gives you flexibility but requires careful management to avoid memory issues.

Pointer Initialization

Pointer initialization is crucial in avoiding undefined behavior in your programs. A pointer variable in C++ holds the memory address of another variable, and when declared, it usually points to a random memory address.
In the given code example, we see `int *p;` and `int *q;` declared without being initialized. These pointers need to be set before use; otherwise, they could lead to accessing memory they shouldn't.

• Using `p = new int;` correctly initializes `p` by setting it to a newly allocated space.
• The line `q = p;` doesn't initialize `q` with a new address, but rather makes it point to the same location as `p`.

Proper initialization ensures that your pointers reference valid memory locations, preventing runtime errors and undefined behaviors.

Memory Address Management

Memory address management involves the careful handling of pointers and their allocated memory to ensure optimal program behavior.
Two aspects were addressed in the solution provided: sharing addresses and updating values. First, with `q = p;`, `q` begins to point to the same address as `p`. This kind of memory address sharing is common when you want multiple pointers to reflect the same data changes.
Then, when `*p = 46;` and subsequently `*q = 39;` are used, it shows how changes to either pointer affect the shared memory location. This is because both `*p` and `*q` are simply different references to the same underlying data.

• Any modifications to either result in the same effect since the memory address remains unchanged.

Efficient memory address management allows for robust and error-free code, making dynamic operations more predictable.