URGENT

Question

the cdde is:

#include <stdio.h>
int sumArray(int [], int); // function prototype

int main(void) {
    int foo[8] = {44, 9, 17, 1, -4, 22};
    int bar[] = {2, 8, 6};
    printf("sum is %d\n", sumArray(foo, 8));
    printf("sum is %d\n", sumArray(foo, 3));
    printf("sum is %d\n", sumArray(bar, 3));
    return 0;
}


int sumArray(int arr[], int size) {
    int i, total=0;
    for (i=0; i<size; i++)
        total += arr[i];
    return total;
}

and outputs are: sum is 89 sum is 70 sum is 16

I cant understand from where these outputs came -_- can anyone please explain?

Answer 1 · 2014-08-18T20:35:34Z

August 18, 2014 8:35pm

foo is an Array of integers up to a max-length of 8. There are 6.

bar is an Array of integers of any length. There are 3.

The second parameter of sumArray is the total amount of items (integers) to use from the given Array in the first parameter.

printf("sum is %d\n", sumArray(foo, 8)); -- This adds up the numbers in foo using up to 8 numbers in the Array. Again, there is only 6 in the array so it's just adds zero twice to equal 8.

printf("sum is %d\n", sumArray(foo, 3)); -- This adds up the numbers in foo using the first 3 items in the Array.

printf("sum is %d\n", sumArray(bar, 3)); -- This adds up the numbers in bar using up to 3 numbers in the Array.

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Kimberly Hedger

Kimberly Hedger

URGENT

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Sean T. Unwin

Sean T. Unwin