c-通过函数放置后输出未更改

Question

This is an extension of a homework question I submitted last night. The program got too messy for me so I took a step back and am starting over with some new knowledge of returns and pass bys. The entire program is meant to take any number of integers, and first find the sum and average, then print the square root of each integer, then print a boolean true or false if any integer is less than 20 and finally print a boolean true or false if any integer is between 10 and 90.

But right now all I need to do is to get my head wrapped around this. Currently I am only going to tackle the sum and average portion of this. It must be done using pass by references which, to my knowledge, is what I have been using in the code shown below:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int functionSumAvg(int sumAvgInput);

void main()
{
  int amountOfNumbers,number,x,output;

  output = 0;

  printf("Enter how many numbers you want: ");
  scanf("%d",&amountOfNumbers);
  printf("Input each number, separated by a space: ");

  for(x = 0; x < amountOfNumbers; ++x){
    scanf("%d",&number);
    int output = functionSumAvg(number);

  }
  printf("output: %d",output);
}


int functionSumAvg(int sumAvgInput){
  int sum =+ sumAvgInput;
  return sum;
}

The idea that I was getting is that you put in the amount of integers you want, then it goes through summing one, adding that sum to the output, and repeating until it's all done. Afterwards it would print out the final output, the result of all the integers summed up. But instead it prints zero. Originally it printed out what I assumed was the memory address, some big trash number, so I set it to zero. But that was only a band aid solution, and didn't solve the problem itself.

Answer 1

   int output = functionSumAvg(number);

creates a new temporary output

do

 output = functionSumAvg(number);

Answer 2

Your problem, is that you are not passing by reference. You are passing by value.

You need to pass-in a reference (a pointer in C) to the sum, and pass-in (by value) the new value to be added to the sum. The calling routine would then be responsible for initialising the sum to zero.

int functionSumAvg(int *sum_pt, int input){…}

I won't tell you the rest, because it will get in the way of your learning.

Note: I use a _pt post-fix for my pointers, I don't use any other Hungarian notation.

Answer 3

fuctionSumAvg doesn't compute the sum or average of anything. It just sets the local variable sum to the number that it received, and returns that same number. It's not adding it to a previous total. And calculating an average requires dividing the sum by the count of elements, but you never sum or count anything.

You can use static variables to hold the running sum and count. But since a function can only return one value, you'll need to use a pointer parameter to return a second value.

int functionSumAvg(int sumAvgInput, int *avg){ // return value is sum
  static int sum = 0;
  static int count = 0;
  sum += sumAvgInput;
  count++;
  *avg = sum/count;
  return sum;
}

Then you call it like this:

int sum, average;
sum = functionSumAvg(number, &average);
printf("Sum = %d Average = %d\n", sum, average);