Compute sum of the array elements using pointers

**DustBin** · 05-25-2016

The stated problem is "compute sum of the array elements using pointers."

The website solution looks like this:

Code:

#include<stdio.h>
#include<conio.h>

void main(){

int numArray[10];
int i, sum = 0;
int *ptr;
 
printf("\nEnter 10 elements : ");
 
for (i = 0;i < 10;i++)
scanf("%d", &numArray[i]);
 
ptr = numArray;/* a=&a[0] */
 
for (i = 0;i < 10;i++){
sum = sum + *ptr;
ptr++;
}
 
printf("The sum of array elements : %d",sum);
}

So instead of
sum += numArray[i]
you have
sum += *ptr;

Ok I get this. Can this also be solved using arrays passed through a function? I.E.

Code:

#include <stdio.h>

void arrayFunc(int myArray...etc); //prototype a function with array and any other variables needed in function

int main() {

int myArray[5]
int sum = 0;
int i;


for (i = 0; i < 5; i++){
printf("Please enter number %d: ", i+1);
scanf("%d", &myArray[i]);
}

for (i = 0; i < 5; i++)
arrayFunc(myArray[i],....)

return 0;
}

void arrayFunc(int myArray, etc..) //other variables may be pointers? i.e. int *sum or something along those lines

//somewhere in here do the sum += myArray[i] using pointers to fill-in-the-blank values

**Salem** · 05-25-2016

> The website solution looks like this:
Which website - not this website surely?

Using conio.h and void main are big red flags to look elsewhere, because the author is stuck in the 1980's.

> void arrayFunc(int myArray...etc); //prototype a function with array and any other variables needed in function
Any standard text would have described how to pass an array to a function.
Look at how you might use say strcat() - it receives arrays of chars.

**rstanley** · 05-25-2016

Instead of:

Code:

void arrayFunc(int myArray, etc..) //other variables may be pointers? i.e. int *sum or something along those lines

You need to pass the array name as a pointer to an int, and the dimension of the array. You also should not use "Magic Numbers" in your code.

The function should be defined as:

Code:

void arrayFunc(int *arrayPtr, int dim)
{
    // ...
}

And call the function as:

Code:

#define DIM 5
// ...
arrayFunc(myArray, DIM);

The name of the array is the address (Pointer) of the first element of the array.

I leave the rest of the code to you. You have everything you need.

If your instructor / book / YouTube video where you learned C did not provide you with this information, then you need a new teaching source!

**DustBin** · 05-25-2016

Originally Posted by rstanley

Instead of:

Code:

void arrayFunc(int myArray, etc..) //other variables may be pointers? i.e. int *sum or something along those lines

You need to pass the array name as a pointer to an int, and the dimension of the array. You also should not use "Magic Numbers" in your code.

The function should be defined as:

Code:

void arrayFunc(int *arrayPtr, int dim)
{
    // ...
}

And call the function as:

Code:

#define DIM 5
// ...
arrayFunc(myArray, DIM);

The name of the array is the address (Pointer) of the first element of the array.

I leave the rest of the code to you. You have everything you need.

If your instructor / book / YouTube video where you learned C did not provide you with this information, then you need a new teaching source!

Thanks for this.

I came up with two versions, messing around with stuff.

Code:

#include <stdio.h>
#include <stdlib.h>
#define SIZE 5;

void arrayFunc(int arr[], int *sum, int i);

int main()
{
   int myArray[SIZE];
   int sum = 0;
   int i;

for (i = 0; i < SIZE; i++)
{
   printf("Please enter #%d: ", i+1);
   scanf("%d'", &myArray[i]);
}

for (i = 0; i < SIZE; i++)
     arrayFunc(myArray, &sum, i);
printf("%d", sum);

return 0;
}

void arrayFunc(int arr[], int *sum, int i)
{
   *sum += arr[i];
}

And keeping in mind your example of intializing as a pointer in the function..

Code:

#include <stdio.h>
#include <stdlib.h>
#define SIZE 5;

void arrayFunc(int *arr, int *sum, int i);

int main()
{
   int myArray[SIZE];
   int sum = 0;
   int i;

for (i = 0; i < SIZE; i++)
{
   printf("Please enter #%d: ", i+1);
   scanf("%d'", &myArray[i]);
}

for (i = 0; i < SIZE; i++)
   arrayFunc(myArray, &sum, i);
printf("%d", sum);

return 0;
}

void arrayFunc(int *arr, int *sum, int i)
{ 
   *sum += *(arr+i);
}

I haven't tried putting the dimensions of the array in the actual function but I will try that too

**rstanley** · 05-25-2016

Code:

void arrayFunc(int arr[], int *sum, int i)
{
   *sum += arr[i];
}

Why do you call this function i number of times? Wouldn't it be more efficient to put the loop in this function, rather than in main()?

Pass i as the NUMBER of the elements of the array, NOT as a single index of ONE element!

I would also return the sum rather than pass a pointer to the sum.

Code:

int arrayFunc(int arr[], int i)
{ 
    int sum = 0;
    // ...
    return sum;
}

**laserlight** · 05-25-2016

Incidentally, going back to the original formulation of the problem, the solution that you were presented boils down to:

Code:

ptr = numArray;
for (i = 0; i < 10; i++) {
    sum = sum + *ptr;
    ptr++;
}

Here the loop index i functions purely as a counter to get the loop to loop as many times as there are elements in the array. This is fine, but then it would have been simpler to write:

Code:

for (i = 0; i < 10; i++) {
    sum = sum + numArray[i];
}

But of course this then fails the "using pointers" part of the challenge. Rather, I suggest that you make use of the pointers when iterating, e.g.,

Code:

for (ptr = numArray; ptr < numArray + 10; ptr++) {
    sum += *ptr;
}

**DustBin** · 05-25-2016

[QUOTE=rstanley;1255826]

Code:

void arrayFunc(int arr[], int *sum, int i)
{
   *sum += arr[i];
}

Why do you call this function i number of times? Wouldn't it be more efficient to put the loop in this function, rather than in main()?

Pass i as the NUMBER of the elements of the array, NOT as a single index of ONE element!

You mean instead of:

Code:

for (i = 0; i < SIZE; i++)
{
   printf("Please enter #%d: ", i+1);
   scanf("%d'", &myArray[i]);
}
 
for (i = 0; i < SIZE; i++)
   arrayFunc(myArray, &sum, i);
printf("%d", sum);
 
return 0;
}
 
void arrayFunc(int *arr, int *sum, int i)
{ 
   *sum += *(arr+i);
}

do something like this(treating the pointers as arbitrary and focusing the proper usage of i)...

Code:

for (i = 0; i < SIZE; i++)
{
   printf("Please enter #%d: ", i+1);
   scanf("%d'", &myArray[i]);
}
 
   arrayFunc(myArray, &sum);
   printf("%d", sum);
 
return 0;
}
 
void arrayFunc(int *arr, int *sum)
{ 
   int i; 
   for (i = 0; i < SIZE; i++)
     *sum += *(arr+i);
}

...?

also, thanks laserlight for your post.

**laserlight** · 05-25-2016

Maybe it would be helpful to see an example:

Code:

#include <stdio.h>

void printNumbers(const int *numbers, int n);

int main(void)
{
    int xs[] = {1, 2, 3, 4, 5};
    int xs_n = 5;

    int ys[10] = {7, 8, 9};
    int ys_n = 3;

    printNumbers(xs, xs_n);
    printf("\n");

    printNumbers(ys, ys_n);
    printf("\n");

    return 0;
}

void printNumbers(const int *numbers, int n)
{
    const int *p;
    for (p = numbers; p < numbers + n; p++)
    {
        printf("%d ", *p);
    }
}

In the above, I am printing numbers, not computing their sum, but the general principles remain. Notice that because I use the parameter n in the loop for printNumbers, printNumbers can be used for both xs and ys, even though they are of different sizes. Not only are they of different sizes, but not all the elements of ys are in use.

Notice that the loop index, or in my case the pointer used to iterate over the array, is not a parameter: it doesn't need to be because it is implementation detail. The caller basically is saying "print these numbers" and passes a pointer to the first element of the array along with the number of elements to print.

**whiteflags** · 05-25-2016

You mean instead of: ...
do something like this(treating the pointers as arbitrary and focusing the proper usage of i)...

No, I think he meant what he said, use i (or pick a better name than i) to mean the number of elements. He gave a partial implementation as an example in his post.

Incidentally, I encourage playing around with things. Here's an alternate implementation that will let you add up any part of an array, or a whole one. It also uses pointers.

Code:

int SumOfSubArray(int *arr, size_t first, size_t last)
{
   int sum = 0;
   int *it;
   int *end;
   /* sum [first .. last] inclusive */
   for (it = arr + first, end = arr + last + 1; it < end; it++) {
      sum += *it;
   }
   return sum;
}

int SumOfArray(int *arr, size_t n)
{
   return SumOfSubArray(arr, 0, n - 1);
}

**DustBin** · 05-26-2016

Originally Posted by whiteflags

No, I think he meant what he said, use i (or pick a better name than i) to mean the number of elements. He gave a partial implementation as an example in his post.

Incidentally, I encourage playing around with things. Here's an alternate implementation that will let you add up any part of an array, or a whole one. It also uses pointers.

Code:

int SumOfSubArray(int *arr, size_t first, size_t last)
{
   int sum = 0;
   int *it;
   int *end;
   /* sum [first .. last] inclusive */
   for (it = arr + first, end = arr + last + 1; it < end; it++) {
      sum += *it;
   }
   return sum;
}

int SumOfArray(int *arr, size_t n)
{
   return SumOfSubArray(arr, 0, n - 1);
}

ok great. Thank you and everyone else for the input =)

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