Here is a simple program in C that meets your specifications: ```c #include void cube(int *n) { *n = (*n) * (*n) * (*n); } int main() { int n; printf("Enter a number: "); scanf("%d", &n); cube(&n); printf("The cube of the number is: %d ", n); return 0; } ``` This program works as follows: 1. It includes the standard input/output library. 2. It defines a function `cube` that takes a pointer to an integer as an argument. It calculates the cube of the number by dereferencing the pointer (getting the value it points to) and multiplying it by itself twice. It then stores the result back in the memory location pointed to by the argument. 3. In the `main` function, it declares an integer `n`. 4. It prompts the user to enter a number and reads the input into `n`. 5. It calls the `cube` function with the address of `n` (which is what the `&` operator does). This is how it passes the number by reference. 6. Finally, it prints the result.

Question

Here is a simple program in C that meets your specifications:

```c
#include <stdio.h>

void cube(int *n) {
    *n = (*n) * (*n) * (*n);
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", &n);
    cube(&n);
    printf("The cube of the number is: %d
", n);
    return 0;
}
```

This program works as follows:

1. It includes the standard input/output library.
2. It defines a function `cube` that takes a pointer to an integer as an argument. It calculates the cube of the number by dereferencing the pointer (getting the value it points to) and multiplying it by itself twice. It then stores the result back in the memory location pointed to by the argument.
3. In the `main` function, it declares an integer `n`.
4. It prompts the user to enter a number and reads the input into `n`.
5. It calls the `cube` function with the address of `n` (which is what the `&` operator does). This is how it passes the number by reference.
6. Finally, it prints the result.

Knowee AI · Accepted Answer

Here is a simple program in C that meets your specifications:

```c
#include <stdio.h>

void cube(int *n) {
    *n = (*n) * (*n) * (*n);
}

int main() {
    int n;
    printf("Enter a number: ");
    scanf("%d", &n);
    cube(&n);
    printf("The cube of the number is: %d
", n);
    return 0;
}
```

This program works as follows:

1. It includes the standard input/output library.
2. It defines a function `cube` that takes a pointer to an integer as an argument. It calculates the cube of the number by dereferencing the pointer (getting the value it points to) and multiplying it by itself twice. It then stores the result back in the memory location pointed to by the argument.
3. In the `main` function, it declares an integer `n`.
4. It prompts the user to enter a number and reads the input into `n`.
5. It calls the `cube` function with the address of `n` (which is what the `&` operator does). This is how it passes the number by reference.
6. Finally, it prints the result.

Question

Solution

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