Chapter 9: Problem 28

Look at the following function definition. void getNumber(int &n) { cout << "Enter a number: "; cin >> n; } In this function, the parameter n is a reference variable. Rewrite the function so that n is a pointer.

Short Answer

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**Answer**: To rewrite the given `getNumber()` function with a pointer, change the parameter type from `int &n` to `int *n`. Keep the same prompt asking the user to enter a number, and modify the `cin` statement to use the dereference operator (*) to access the value stored at the memory location pointed by `n`. The updated function should look like this: ```cpp void getNumber(int *n) { cout << "Enter a number: "; cin >> *n; } ```

Step by step solution

Change the parameter type to a pointer

Change the parameter type in the function definition from `int &n` to `int *n`. ```cpp void getNumber(int *n) ```

Modify the prompt to ask for pointer input

In the function body, keep the same prompt asking the user to enter a number. This will remain unchanged because the user doesn't need to know whether we are using a reference or a pointer in the function. ```cpp cout << "Enter a number: "; ```

Update the `cin` statement to use a pointer

Change the `cin` statement to use the pointer instead of the reference. We should use the dereference operator (*) to access the value stored at the memory location pointed by `n`. ```cpp cin >> *n; ``` After making these changes, our final function will look like this: ```cpp void getNumber(int *n) { cout << "Enter a number: "; cin >> *n; } ``` This is the updated version of the function using pointer instead of reference variable.

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Look at the following function definition. void getNumber(int &n) { cout << "Enter a number: "; cin >> n; } In this function, the parameter n is a reference variable. Rewrite the function so that n is a pointer.

Short Answer

Step by step solution

Change the parameter type to a pointer

Modify the prompt to ask for pointer input

Update the `cin` statement to use a pointer

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