Function to count the sum of an array - c

Question

Right, this is (the last) assignment for my C introduction web class.
The assignment presents the main program, does not explain anything about it and tells you to write a function to print and sum the array in it.
However I don't really understand what is going on in the main program.
Translated for your convenience;
Source code:
#include <stdio.h>
#include <stlib.h>
void print_count(int *, int);
int main(int argc, char *argv[]) {
int x, sum = 0, size = 0, array[5];
if (argc == 6) {
/* Program name and parameters received from command line */
for (x = 0; x < argc - 1; x++) {
array[x] = atoi(argv[x + 1]);
}
print_count(array, size);
} else {
printf("Error\n");
}
return 0;
}
Now I am completely clueless as to how to start writing the program requested and what variables to call/how to write the function.
Edit3: completed exercise
void print_count(int *array, int size) {
int i;
int sum = 0;
printf("Elements: ");
for (i = 0; i <= size; i++) {
printf("%d ", (array[i]);
sum = sum += array[i]);
}
printf("\nSum = %d ", sum);
return 0;
}
I would like to understand what is going on in the main program and preferably come to an answer on how to actually write the function by myself.

Answer 1

This:
array[5] = atoi(argv[x+1]);
is clearly wrong, it always tries to assign to array[5] which is out of bounds. It should be:
array[x] = atoi(argv[x + 1]);
This converts the x + 1:th argument from string format into an integer, and stores that in array[x]. If you're not familiar with the standard function atoi(), just read the manual page.
So if you start the program like this:
./myprogram 1 2 3 4 5
That has 6 arguments (the first is the name itself), and will end up with array containing the numbers one through five.
Then in the summing function, the first line should be something like:
void print_count(int *array, int size)
so that you give names to the arguments, which makes them usable in the function. Not providing names is an error, I think.
And it doesn't need to "interact" with main() more than it already does; main() calls print_count(), passing it a pointer to the first element of array and the length of the array, that's all that's needed to compute the sum.

Answer 2

Your print_count function has a few issues:
The loop runs one step too far: i should vary between 0 and size-1 included. The standard idiom for this loop is:
for (i = 0; i < size; i++) {
...
Incrementing sum is simply done with:
sum += array[i];
There is an extra ( on the first printf line.
You should print a newline after the output.
Returning 0 from a void function is invalid.
Here is a corrected version:
void print_count(int *array, int size) {
int i;
int sum = 0;
printf("Elements: ");
for (i = 0; i < size; i++) {
printf("%d ", array[i]);
sum += array[i]);
}
printf("\nSum = %d\n", sum);
}

Answer 3

the following proposed code:
cleanly compiles.
explains what is being accomplished at each step of the 'main()' function.
properly outputs error messages to 'stderr'.
implements the typical method to announce an error in the number of command line parameters.
Now the proposed code with explanatory comments:
#include <stdio.h> // printf(), fprintf()
#include <stdlib.h> // atoi(), exit(), EXIT_FAILURE
void print_count(int *, int);
int main(int argc, char *argv[])
{
if (argc != 6)
{
fprintf( stderr, "USAGE: %s int1 int2 int3 int4 int5\n", argv[0] );
exit( EXIT_FAILURE );
}
// implied else, correct number of arguments
// only declare variables when they are needed
int array[5];
// place each command line parameter into 'array',
// except program name
// I.E. skip the program name in argv[0]
for( int i = 1; i < argc; i++ )
{
// array[] index starts at 0, but loop counter starts at 1
array[i-1] = atoi(argv[i]);
} // end for( each value pointed at by argv[], except program name )
// print sum of command line parameters to stdout
int size = argc-1; // number of command line parameters after program name
print_count(array, size);
return 0;
} // end function: main