C function with variable number of arguments returns unexpected output when called with fewer number of arguments

Question

I'm learning C, and I created a simple addNumbers(int num, ...) function, that takes any number of int arguments, and returns the sum of them.

This issue is that when I get the output of addNumbers(4, 2), the output is 14823901. Which is obviously incorrect.

When I call the function addNumbers(4, 2, 4, 7, 10), it outputs 23, which is also incorrect because it should be 27, but at least it's closer.

Here's my code:

#include<stdio.h>
#include<stdarg.h>

// Functions with variable number of arguments
int addNumbers(int num, ...)
{
    int i;
    int sum = 0;

    // List to hold variable amount of parameters
    va_list parameters;

    // Initialize "parameters" list with arguments
    va_start(parameters, num);

    for(i = 0; i < num; i++)
    {
        // Adds each "int" argument from "parameters" to sum
        sum += va_arg(parameters, int);
    }

    // Cleans memory
    va_end(parameters);

    return sum;
}

int main()
{
    printf("%i", addNumbers(4, 2, 4, 7, 10));
    return 0;
}

Should I not be using va_list, va_arg, etc...?

What's the best way to be able to take in a variable number of arguments?

Answer 1

for addNumber(4, 2) you are using the first parameter as counter which there are 4 parameter to addup but your giving just 1, so the for loop continue reading from the memory expecting more parameter and just pick up ramdom values and add them up.