Keeping geting a zero in the main function - c

Question

I added a few printf stantments to see if the information is being computer and passed correctly, and while is is being computed correctly in the calcMean function, I keep getting zeros in the main function. I am not sure where my mistake is.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void calcMean(int [], int, float);
void calcVariance(int [], int, float);
int main(void)
{
/*create array and initialize it with values*/
int mainArr [ ] = {71,1899,272,1694,1697,296,722,12,2726,1899,
.......
1652,488,1123,17,290,1324,2495,1221,2361,1244,
813,2716,1808,2328,2840,1059,2382,2391,2453,1672,
1469,778,2639,357,2691,1113,2131,23,2535,1514,
2317,45,1465,1799,2642,557,1846,1824,1144,1468,-1};
/*create initilized values for variables*/
float mean = 0.0;
int counter = 0;
calcMean(mainArr, counter, mean);
calcVariance(mainArr, counter, mean);
printf ("Mean: %10.2f\n", mean); /*gives the average number to one decimal place*/
printf ("counter: %10.0d\n", counter);
getchar();
}/*End of the program*/
void calcMean(int mainArr[], int counter, float mean)/*This function finds the Mean(average) of the function, and the size of the array*/
{
int sentinel = -1;
float sum = 0.0;
int index = 0;
for (index; mainArr[index] != sentinel; index++) /*gets the total numbers in the function and the sum*/
{
counter++;
sum = sum + mainArr[index];
} /*end of the sumation and counting of integers*/
printf ("%0.0d\n", counter);
mean = ((float) sum / counter); /*Divides the sum by the total number of numbers to find the mean*/
printf ("mean: %10.2f\n", mean);
}
void calcVariance(int mainArr[], int counter, float mean)
{
int varindex = 0;
int numerator =0;
int square = 0;
int variance = 0;
int sumation = 0;
for(varindex; varindex<counter; varindex++)
{
numerator = (mainArr[varindex] - mean);
square = (numerator * numerator);
sumation = sumation + square;
variance = sumation / (counter-1);
}
printf ("variance %10.2f\n", variance);
}

Answer 1

You're assuming your function changes the value of outer variables. It won't.
calcMean(mainArr, counter, mean);
^ ^ pass by value
void calcMean(int mainArr[], int counter, float mean)
^ ^ receive by value
Modifying counter inside the function changes only the local variable.

Answer 2

Function parameters are copied into functions. So the variables inside a function will not change the parameters in the outer function.
You will need to do something like the following to any function which you wish to pass back more than one value from a function. (If you only need to pass back one value that can be returned with the function return statement).
// declare the parameters you wish to pass back to calling function as pointers
void calcMean(int [], *int, *float);
// pass the address of the variables:
calcMean(mainArr, &counter, &mean);
// obtain the values of the pointers in your calcMean using derefence operator *
printf ("%0.0d\n", *counter);
*mean = ((float) sum / *counter);

Answer 3

OP later requested an answer with little use of pointers.
As others (#suspectus, #Karoly Horvath, #Carl Norum) correctly pointed out the problem that passing into a function mean does not change it.
main() has an object named mean. That object's value was used to initialize the object called mean in the argument list of calcMean(). calcMean() proceeded to set its mean to some value. At the end of the function, the original mean in main() is not affected.
Instead, use the return value of function calcMean() to set mean in main().
// void calcMean(int mainArr[], int counter, float mean)
float calcMean(int mainArr[], int counter) {
float mean;
...
mean = ((float) sum / counter);
return mean;
}
int main(void) {
...
float mean = 0.0;
...
// calcMean(mainArr, counter, mean);
mean = calcMean(mainArr, counter);
...
}
Do the same for calcVariance

Answer 4

C is a pass-by-value language. Modifying mean and counter in the other functions doesn't affect their values in main().