Calculate the sum of two numbers and "return" it by using a void function in C with a pointer as parameter

Question

So basically I'd like to sum two numbers and return their value while using a void function in C. I know this is easy peasy by using a normal function returning an int or a numeric type but I wanna work on my pointer knowledge.

I tried creating a pointer inside main() and then passing it as an argument to the void function. Then I calculated my sum in a new int variable and assigned the pointer to point to that specific variable. The problem is I can't "retrieve" it or "find" that area of memory in the main function.

Here's what I've tried:

void testFunction(int a,int b, int *x)
{
   int c=a+b;
   x=&c; 
}

int main()
{
  int n1=7;
  int n2=90;
  int *pointerParam;
  testFunction(n1, n2, pointerParam);
   
  printf("Value of pointer is %d\n", *pointerParam);
}

It just exits with an error code, it does nothing. If I try to printf *x inside the function, it does work so I know that part at least works.

Any help would be greatly appreciated!

Answer 1

You lacked just to understand how pointers are managed, but you were almost correct:

/* this is almost correct, you could have just said: *x = a + b; */
void testFunction(int a,int b, int *x)
{
   int c=a+b;
   *x=c; /* the pointed to value is what we are assigning */
}

int main()
{
  int n1=7;
  int n2=90;
  int result;       /* vvvvvvv this is the important point */
  testFunction(n1, n2, &result); /* you pass the address of result as the required pointer */
   
  printf("Value of pointer is %d\n", result);
}

Answer 2

I will show you a program that is no more than a set of printf() and your function. Maybe the program's output helps in showing these pointers, arrays and integers things.

The example

Consider these variables

    int  n[4]      = {10, 20, -30, -40};
    int  v1        = 0;
    int  v2        = 0;
    int* a_pointer = NULL;

int* is a pointer to an int, that's the meaning of the asterisk in the declaration. In this context the asterisk is called the dereference operator or the indirection operator. But the asterisk also is the multiplication operator in C. ;)

Now a_pointer points to nothing, the meaning of the NULL.

But a_pointer is there to hold an address of something, of an int. The way of getting such address of something is the address of operator, the &, that also has an alternate life as the bitwise and operator in C. Things of life.

In printf() the %p specifier shows an address. This

    printf("\naddress of v1 is %p\n", &v1);
    printf("address of v2 is %p\n", &v2);
    printf("address of array n[0] is %p\n", n);
    printf("address of array n[0] is %p\n", 1 + n);
    printf("address of array n[0] is %p\n", 2 + n);
    printf("address of array n[0] is %p\n\n", 3 + n);

shows (in a 32-bits compilation)

address of v1 is 008FFDA0
address of v2 is 008FFD9C
address of array n[0] is 008FFDA4
address of array n[1] is 008FFDA8
address of array n[2] is 008FFDAC
address of array n[3] is 008FFDB0

And you will see the reason the program prints these lines in the code below...

This line

    a_pointer = &v1;

takes the address of v1 and assign it to the pointer a_pointer. Now a_pointer is pointing to something, to v1, and you can use it in your function. These lines are equivalent

    testFunction(n[0], n[3], &v1);

and

    testFunction(n[0], n[3], a_pointer);

In these lines

    testFunction(n[0], n[3], &v1);
    printf("n[0] = %d, n[3] = %d, sum in v1 is %d\n", n[0], n[3], v1);
    printf(
        "p points to v1. value is %4d, address is %p\n\n", *a_pointer,
        a_pointer);

you see the use of the pointer to access the value it points to, using the dereference operator in the printf().

Follow the program along to see a few uses of this.

In particular, see these lines

    a_pointer = n + 3;
    printf(
        "\np points now to n[3]. value is %4d, address is %p\n",
        *a_pointer, a_pointer);

to see the thing C is made for: address memory easily. n is int[4], an array of int. a_pointer is a pointer to int. And the language knows that when you write a_pointer = n + 3 that it needs to add to the address of n, an int[], the size of 3 int variables, and assign it to the pointer, so *a_pointer is n[3]and it is used to call testFunction() in

    testFunction(n[1], n[2], n + 3);

program output

n[] is [10,20,-30,-40], v1 is 0 v2 is 0

address of v1 is 00CFFA9C
address of v2 is 00CFFA98
address of array n[0] is 00CFFAA0
address of array n[1] is 00CFFAA4
address of array n[2] is 00CFFAA8
address of array n[3] is 00CFFAAC

n[0] = 10, n[3] = -40, sum in v1 is -30
p points to v1. value is  -30, address is 00CFFA9C

p now points to v2. value is    0, address is 00CFFA98
n[0] = 10, n[1] = 20, sum in v2 is 30
n[] is [10,20,-30,-40], v1 is -30 v2 is 30

p now points to v1. value is  -30, address is 00CFFA9C
n[] is [10,20,-30,-40], v1 is -30 v2 is 30

now makes n[3] = n[1] + n[2] using testFunction()
n[] is [10,20,-30,-10], v1 is -30 v2 is 30

p points now to n[3]. value is  -10, address is 00CFFAAC

the code

#include <stdio.h>

show(int[4], int, int);
void testFunction(int, int, int*);

int main(void)
{
    int  n[4]      = {10, 20, -30, -40};
    int  v1        = 0;
    int  v2        = 0;
    int* a_pointer = NULL;

    show(n, v1, v2);
    printf("\naddress of v1 is %p\n", &v1);
    printf("address of v2 is %p\n", &v2);
    printf("address of array n[0] is %p\n", n);
    printf("address of array n[1] is %p\n", 1 + n);
    printf("address of array n[2] is %p\n", 2 + n);
    printf("address of array n[3] is %p\n\n", 3 + n);
    a_pointer = &v1;
    testFunction(n[0], n[3], &v1);
    printf("n[0] = %d, n[3] = %d, sum in v1 is %d\n", n[0], n[3], v1);
    printf(
        "p points to v1. value is %4d, address is %p\n\n", *a_pointer,
        a_pointer);

    a_pointer = &v2;
    printf(
        "p now points to v2. value is %4d, address is %p\n", *a_pointer,
        a_pointer);
    testFunction(n[0], n[1], &v2);
    printf("n[0] = %d, n[1] = %d, sum in v2 is %d\n", n[0], n[1], v2);
    show(n, v1, v2);

    a_pointer = &v1;
    printf(
        "\np now points to v1. value is %4d, address is %p\n", *a_pointer,
        a_pointer);

    show(n, v1, v2);
    printf("\nnow makes n[3] = n[1] + n[2] using testFunction()\n");
    testFunction(n[1], n[2], n + 3);
    show(n, v1, v2);

    a_pointer = n + 3;
    printf(
        "\np points now to n[3]. value is %4d, address is %p\n",
        *a_pointer, a_pointer);

    return 0;
};

show(int n[4], int v1, int v2)
{
    printf(
        "n[] is [%d,%d,%d,%d], v1 is %d v2 is %d\n", n[0], n[1], n[2],
        n[3], v1, v2);
};

void testFunction(int a, int b, int* sum)
{
    *sum = a + b;
    return;
}

I will not go into religious discussions here, but you may find easier to understand the meaning of the declarions if you write

    int*    some_int = NULL;

instead of

    int    *some_int = NULL;

you declare a name, and the name is some_int. The compiler will tell you that some_int is int*, its type. The fact that *some_int is an int is a consequence of the application of an operator to a variable.

Answer 3

There are multiple problems with the code as it is shown.

The main problem is probably that you misunderstand how emulation of pass-by-reference works in C.

For it to work you need to pass a pointer to the variable that should be set. This is done using the pointer-to operator &. You also need to dereference the pointer, to set the variable the pointer is pointing to.

Putting it together your program should look something like this (simplified):

void testFunction(int a,int b, int *x)
{
    // Assign to where `x` is pointing
    *x = a + b;
}

int main(void)
{
    int n1 = 7;
    int n2 = 90;
    int result;  // Where the result should be written

    // Pass a pointer to the `result` variable, so the function can write to it
    testFunction(n1, n2, &result);
}