Solution for Exercise 03_04-added.c

03_04-added.c

#include <stdio.h>

void added(int *a, int *b)
{
	*b = *a * 2;
}

int main()
{
	int x,y;

	x = 22;
	added(&x,&y);
	printf("%d doubled is %d\n",x,y);

	return 0;
}

Output

22 doubled is 44

Notes

* Obviously, your solution need not look exactly like what's shown above.

* The added() function requires two pointer arguments, which I define as a and b.

* The value of a (written as *a) is doubled: *a * 2

* The compiler is smart enough to know the difference between the * pointer operator, which binds tightly to variable a, and the * multiplication operator.

* The result of *a * 2 is stored at the address in pointer variable b.

* No value needs to be returned from the added() function as the value generated is stored directly in memory and accessed from the main() function.

* In the main() function, I use integer variables x and y. Variable x is assigned the value 22: x = 22;

* The addresses of variables x and y are passed to the added() function: added(&x,&y). The & operator satisfies the function's desire for pointers/addresses.

* Variable y isn't initialized when it's passed to the added() function. This condition is unacceptable for a pointer variable, but for a non-pointer variable like y it's tolerable. Because I know that the added() function doesn't use any garbage value already stored in memory, it's okay. Always be aware when passing uninitialized variables to a function.

Exercise 03_07-both.c →