summing numbers using pointer arithmetic - c

Question

I was given an assignment to write a code which takes in numbers as input from the user and provides the sum of it, specifically by the use of pointer arithmetic i.e. no array subscripting a[i] is allowed.
Below is the code that I wrote, which got compiled and even ran. But almost always it gives the sum of the input numbers as 0. I tried to fix it, but to no avail. Thus, I am asking for help, any help is greatly appreciated.
#include<stdio.h>
#define N 5
int sum_array( const int *p, int n)
{
int sum, a[N];
sum = 0;
for(p=&a[0]; p<&a[N]; p++)
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(i,N);
printf("the sum is %d\n", x);
return 0;
}

Answer 1

Beware, you are declaring array int a[N] in both main and sum_array. They are in different scopes, so they are different arrays (and the one from sum_array is never initialized so reading it invokes Undefined Behaviour).
The correct way is to pass the array along with its used length:
Here is a fixed version:
#include<stdio.h>
#define N 5
int sum_array( const int *a, int n) // a points to a array of at least n elements
{
int sum = 0; // initialize at definition time
for(const int *p=a; p<&a[n]; p++) // have the pointer p take all values from a
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}
Finally it is mainly a matter of taste, but I was too often burnt for trying to add an instruction in a for loop without braces, so I strongly recommend using always braces for loops:
for(i=a; i<a+N; i++) {
scanf("%d", i);
}

Answer 2

int sum_array( const int *p, int n)
{
int sum = 0, i = 0;
for(i = 0; i < n ; i++)
sum += *(p+i);
return sum;
}
int main(void)
{
int a[N], i = 0, x = 0;
printf("Enter %d Numbers: ", N);
for(i=0; i<N; i++)
scanf("%d", a+i);
// all the input values get scanned as i or the array a
x= sum_array(a,N);
printf("the sum is %d\n", x);
return 0;
}

Answer 3

In x= sum_array(i,N); i is the iterator of your loop so after the loop has finished it points to the first position after the array.
You should pass the original array instead x= sum_array(a,N);
In the sum function you just throw away the passed pointer and replace it with your local a[].
int sum_array( const int *p, int n)
{
int sum = 0;
int *end = &p[n]; // first element after the array.
for(; p<end; p++) // just use p because you don't need the reference to the start of the array
{
sum += *p;
}
return sum;
}
but as you said that array notation is not allowed you can change it as follows
#include "stdio.h"
#define N 5
int sum_array( const int *p, int n)
{
int sum = 0;
const int *end = p+n; // first element after the array.
for(; p<end; p++)
{
sum += *p;
}
return sum;
}
int main()
{
int *a, *i, x;
a = malloc(N * sizeof(*a));
if (a == NULL)
exit(-1);
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
{
scanf("%d", i);
}
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}

Answer 4

in general, when programming, the code should be kept as simple as possible while still being complete.
The program criteria shows no need to keep a number after it is applied to the sum of the numbers, So in the proposed code, the input number is only kept long enough to be applied to the sum, then it is discarded.
The following proposed code:
cleanly compiles
performs the desired functionality
is kept very simple
properly checks for; and handles any errors
And now the proposed code:
#include <stdio.h> // scanf(), printf(), fprintf(), stderr
#include <stdlib.h> // exit(), EXIT_FAILURE
#define N 5
int main( void )
{
int num = 0;
int sum = 0;
printf("Enter %d Numbers: ", N);
for(size_t i=0; i<N; i++)
{
if( scanf("%d", &num) != 1 )
{
fprintf( stderr, "failed to read number\n" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
sum += num;
}
printf( "the sum is %d\n", sum );
return 0;
}