递归函数从int数组三乘三元素计算平均值

Question

Calculating average three by three elements and replacing those elements with the average result.

Example array [1,2,7,-2,5,0, 2,8] After transformation [3,3,3,1,1,1,5,5]

Something is wrong, I can't get it to work.

#include <stdio.h>

int main ( )    {
    int n, c[n];
    int *avg;
    int pom=0;

    printf("Enter lenght of array\n");

    scanf("%d",&n);

    printf("Enter elements");

    for(i = 0;i < n; i++)
        scanf("%d",c[i]);

    avg=Average(c , n, pom);

    for(i = 0; i < n; i++)
        printf("Avg elements= %d",*(avg+i))
    return 0;
}

int Average(int arr[], int size, int z)
{
    int k, l, m, Asum;

    if (size < 0) {
        return arr;
    } else {
        k=arr[z];
        l=arr[z+1];
        m=arr[z+2];

        Asum=(k + l + m)/3;

        arr[z]=Asum;
        arr[z+1]=Asum;
        arr[z+2]=Asum;
    }

    return Average(arr,size--,z++);
}

Answer 1

int n, c[n]; is a problem. n is uninitialized so the size of the array is who-knows-what? This is undefined behavior.

Instead

int main(void) {
  int n;
  int *avg;
  int pom=0;

  printf("Enter length of array\n");
  if (scanf("%d",&n) != 1) return -1;
  int c[n];

  for(i = 0;i < n; i++)
    // scanf("%d",c[i]);
    scanf("%d",&c[i]);  // pass the address of an `int`

Likely other issues too.

Answer 2

Try simple input first, imagine what happens when you enter only 1 number, what will the Average function do? Don't run the code but try to execute it in your head or with pencil and paper. If you think the program only has to work with three or more numbers, try three.

A serious program would explicitly reject invalid input.