Calculation of average returns wrong result - c

Question

The for loop I wrote to calculate sum of numbers is not working properly.
The program finds the average of the given numbers using a dynamically allocated array.
#include<stdio.h>
main()
{
int no_of_nos;
printf("Enter the number of numbers yuo want to do average : ");
scanf("%d",&no_of_nos);
int x[no_of_nos+1];
int i;
for(i=0;i<no_of_nos;i+=1)
{
printf("Enter the values for the element of the array : ");
scanf("%d",&x[i]);
}
int sum=0,j;
for(j=0;j<no_of_nos;j++)
{
sum = x[j] + x[j+1];
}
printf("The sum of the given numbers = %d \n",sum);
float average = sum / no_of_nos;
printf("The average of the given numbers = %0.2f",average);
}

Answer 1

Your second loop is wrong, it should read:
for( i=0; i<no_of_nos; i++ ) {
sum += x[i];
}
No need for another integer, 'i' will do fine. You could optimise it further by moving the sum up before the first loop then totalise sum as you enter the values. Do you need the array?
int sum = 0;
for( int i=0; i<no_of_nos; i++ ) {
int x;
printf("Enter the values for the element of the array : ");
scanf("%d", x);
sum += x;
}

Answer 2

For starters according to the C Standard the function main without parameters shall be declared like
int main( void )
There is no sense to declare the array x with no_of_nos+1 elements instead of no_of_nos elements. So the declaration of the array should look like
int x[no_of_nos];
In fact an array is not required if you need only to calculate the average of entered numbers.
It is better initially to declare the variable sum as having type float or double. Otherwise this expression statement
float average = sum / no_of_nos;
does not make a great sense because the expression sum / no_of_nos evaluates as an integer expression that is in any case the result of the expression is an integer.
However if the variable sum declared like for example float then the expression sum / no_of_nos will have the type float.
For example
float sum = 0;
//...
float average = sum / no_of_nos;
Compare two outputs of the average in the following demonstrative program.
#include <stdio.h>
#define N 2
int main(void)
{
int a[N] = { 1, 2 };
int sum1 = 0;
for ( size_t i = 0; i < N; i++ ) sum1 += a[i];
float average = sum1 / N;
printf( "The first average is equal to %0.2f\n", average );
float sum2 = 0;
for ( size_t i = 0; i < N; i++ ) sum2 += a[i];
average = sum2 / N;
printf( "The second average is equal to %0.2f\n", average );
return 0;
}
The program ooutput is
The first average is equal to 1.00
The second average is equal to 1.50
This loop
for(j=0;j<no_of_nos;j++)
{
sum = x[j] + x[j+1];
}
also does not make sense. It is enough to write
for(j=0;j<no_of_nos;j++)
{
sum += x[j];
}

Answer 3

The program to calculate the Average of given numbers can be written without the use of arrays.
The following program finds the average of given numbers and gives the answer upto two decimal places:
#include<stdio.h>
main()
{
int n,i=0,j;
float sum=0;
printf("Enter the number of elements : ");
scanf("%d",&n);
for(;i<n;i++)
{
printf("Enter the number : ");
scanf("%d",&j);
sum = sum + j;
}
printf("The sum of the given numbers = %0.2f \n",sum);
float average = sum / n;
printf("The Average of the given numbers = %0.2f",average);
}