Pointers

Given these lines of code, show what would be printed. Let's address in memory of y: 8086, x: 2048

Answer explanation

• x holds the value 10.

• y is a pointer to x, meaning *y (dereferencing y) gives the value stored at x, which is 10.

Given the following lines of code, show what would be printed. Assume that the address in memory is x: 080, y: 360, and z: 551.

Answer explanation

This is because x is 100, and both pointers y and z dereference to the value of x.

What is wrong with the following line of code?

• The value 25 should be declared as a pointer type.

• The pointer is not initialized correctly; it should point to a valid memory address.

• The pointer x should not be declared as int*.

• There is no error; the code is valid.

Answer explanation

The code attempts to assign an integer value (25) directly to a pointer (int* x). Pointers should be assigned the address of a variable, not an integer value.

What is wrong with the following lines of code?

• The variable x cannot be assigned to a pointer of a different type.

• The const qualifier means x cannot be referenced by a pointer.

• The pointer y should be of type const int* instead of double*.

• There is no error; the code is valid.

Answer explanation

The line of code attempts to assign a pointer of a different type to a variable that is intended to point to a const int, leading to a type mismatch and potential issues with modifying the pointed-to value.

What is wrong with the following lines of code?

• The variable x cannot be assigned to a pointer.

• The pointer y should be of type const int* instead of int*.

• The variable x must be mutable to be pointed to by a non-const pointer.

• There is no error; the code is valid.

Answer explanation

The code is incorrect because x is declared as const int, meaning its value cannot be modified. A non-const pointer (int* y) suggests that the value can be changed, which contradicts the const qualifier. To maintain const correctness, y should be declared as const int*.

What is wrong with the following lines of code?

• You cannot have two variables with the same name in the same scope.

• The pointer y is declared correctly the first time but incorrectly the second time.

• The variable x cannot be assigned to a pointer.

• There is no error; the code is valid.

Answer explanation

The primary issue in the code is the redeclaration of the variable y with a different type in the same scope, which is not allowed in C++.

What is wrong with the following lines of code?

• The variables x and y cannot be assigned to a pointer.

• The pointer z should not be declared as const.

• z cannot be reassigned to point to a different address since it is a constant pointer.

• There is no error; the code is valid.

Answer explanation

The error in the code arises from attempting to reassign a constant pointer, which is not permitted.

What is printed from the following lines of code?

Answer explanation

• Pointer Arithmetic: cout << *(sample + 2);

• The expression sample + 2 points to the address of the third element in the array (sample[2]).

• The * operator dereferences this pointer, which means it retrieves the value at that address.

What is printed from the following lines of code?

Answer explanation

• cout << *sample + 2 << endl;

• *sample dereferences the pointer to the first element of the array, which is sample[0] (value 5).

• The expression *sample + 2 evaluates to 5 + 2, which is 7.

  Therefore, first line prints 7.
  cout << *(sample + 2);
  The expression sample + 2 points to the address of the third element in the array (sample[2]).

• The * operator dereferences this pointer, retrieving the value at that address, which is 15.

• Therefore, this line prints 15.

What is printed from the following lines of code?

• The code will print 0.

• The code will print a garbage value.

• The code will cause a segmentation fault.

• The code will print 40.

Answer explanation

Accessing *(sample + 7) is out of bounds since the sample array only has 5 elements (indices 0 to 4). This leads to undefined behavior, which may result in printing a garbage value, causing a crash, or potentially accessing unintended memory. However, the most common scenario is that it prints a garbage value.