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What is the output of the following C++ code? int *p; int *q; p = new int; q = new int; *p = 27; *q = 35; cout << *p << " " << *q << endl; q = p; *q = 73; cout << *p << " " << *q << endl; p = new int; *p = 36; *q = 42; cout << *p << " " << *q << endl;

Short Answer

Expert verified
Outputs: '27 35', '73 73', '36 42'. Likely undefined due to memory handling.

Step by step solution

01

Variable Initialization

The code begins by declaring two pointer variables, `int *p;` and `int *q;`. At this point, they are uninitialized and contain garbage memory addresses.
02

Memory Allocation

The pointers are then set to point to new integer memory locations with `p = new int;` and `q = new int;`. This allocates memory in the heap for each pointer.
03

Assigning Values to Pointers

The values `27` and `35` are assigned to the locations pointed to by `p` and `q` respectively, using the statements `*p = 27;` and `*q = 35;`.
04

First Output

The first output `cout << *p << " " << *q << endl;` prints the values stored at the locations pointed by `p` and `q`, which are `27` and `35`, respectively.
05

Pointer Reassignment

The line `q = p;` makes `q` point to the same memory location as `p`, thus making it lose its original reference, which creates a memory leak for the previous `new int` assigned to `q`.
06

Modify Shared Memory

After `q = p;`, both `*p` and `*q` refer to the same memory location. The value `73` is assigned with `*q = 73;`, which changes the value at this memory location from `27` to `73`.
07

Second Output

The second output `cout << *p << " " << *q << endl;` will display `73` `73` since both pointers point to the same memory location now containing `73`.
08

New Memory Allocation for p

The statement `p = new int;` allocates a new memory location for `p`, leaving the previous memory block containing `73` still pointed to by `q`. The original `73` value now remains unchanged.
09

Assign New Values

The value `36` is assigned with `*p = 36;`, and `42` is attempted on `*q` (which still points to the last shared location). `*q` therefore now attempts to update to `42`, but it affects original value space before reallocation.
10

Third Output

The last line `cout << *p << " " << *q << endl;` prints `36` and the result of the `42` assignment could cause undefined behavior since we altered a previously moved allocation but `q` targeted there.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pointer Variables
In C++, pointer variables are a powerful feature that allow you to handle memory addresses directly. A pointer variable is declared using an asterisk (*) before the variable name, like `int *p;`. This tells the compiler that `p` is a variable that will hold the address of an `int` type data.

When you declare a pointer, it doesn't automatically point to a valid memory address. Initially, it may contain a garbage address unless it is explicitly initialized to point to a specific location.

Here are some key considerations when using pointers:
  • Always initialize pointers to prevent them from pointing to unintended memory addresses.
  • Use the `&` operator to get the address of a variable, which can then be assigned to a pointer.
  • Access or modify the value at the pointed address using the dereference operator `*`.
Memory Allocation
Memory allocation in C++ is about reserving memory space in either the stack or the heap. Dynamic memory allocation is done using the `new` operator, which allows you to allocate memory at runtime.

For example, `p = new int;` allocates memory dynamically and assigns the address of that memory to `p`. This means `p` now holds the address of a dynamically created integer.

Important points to remember about dynamic memory:
  • Always pair every `new` with a corresponding `delete` to release memory and avoid memory leaks.
  • Dynamic allocation provides more flexibility than static memory allocation.
  • Heap memory is limited; watch out for excessive allocations that can exhaust the resources.
Memory Leak
A memory leak occurs when dynamically allocated memory isn't properly freed, thus becoming unreachable. In the example, when `q = p;` is executed, the originally allocated memory for `q` is lost because `q` reassigns to point to what `p` is pointing to.

To prevent memory leaks, follow these best practices:
  • Always use `delete` or `delete[]` for every `new` or `new[]` you use to allocate memory.
  • Before reassignment of pointers, ensure any dynamically allocated memory is properly deallocated.
  • Consider using smart pointers, like `std::unique_ptr` or `std::shared_ptr`, which automatically manage memory and reduce the risk of leaks.
  • Tools and features such as sanitizers and static analysis can help identify leaks during testing phases.
Pointer Reassignment
Pointer reassignment changes where a pointer is pointing. When you use `q = p;`, you make `q` point to the same memory location as `p`. After reassignment, both `p` and `q` will refer to the same memory.

This can be a useful technique but requires caution:
  • Be aware of what the original pointer (`q` in this case) was pointing to, as it's easy to accidentally create a memory leak if you overwrite the address without first freeing it.
  • Remember that changing the value at the shared memory location will affect all pointers referring to that memory. Always trace all pointers involved.
  • Reassignment does not free the original memory, so ensure proper memory management after such operations.
Pointer reassignment can lead to complex relationships in your code, so handling them carefully is crucial for maintaining data integrity and preventing resource depletion.

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Most popular questions from this chapter

Suppose that you have the following classes, classA and classB: class classA { public: virtual void print() const; void doubleNum(); classA(int a = 0); private: int x; }; void classA::print() const { cout << "ClassA x: " << x << endl; } void classA::doubleNum() { x = 2 * x; } classA::classA(int a) { x = a; } class classB: public classA { public: void print() const; void doubleNum(); classB(int a = 0, int b = 0); private: int y; }; void classB::print() const { classA::print(); cout << "ClassB y: " << y << endl; } void classB::doubleNum() { classA::doubleNum(); y = 2 * y; } classB::classB(int a, int b) : classA(a) { y = b; } What is the output of the following function main? int main() { classA *ptrA; classA objectA(2); classB objectB(3, 5); Exercises | 855 ptrA = &objectA ptrA->doubleNum(); ptrA->print(); cout << endl; ptrA = &objectB ptrA->doubleNum(); ptrA->print(); cout << endl; return 0; }

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