C Program: Array c[i] = a[i] + b[n-1-i]

Given two integer arrays a[] and b[] of n elements, compute a third array c[] where each element is c[i] = a[i] + b[n-1-i]. The key insight is the index n-1-i: when i=0 it selects the last element of b, when i=n-1 it selects the first. This “reverse pairing” adds each element of a to the corresponding element of b read in reverse order. The original post contained only a dead Gist shortcode with no rendered code; this is a complete C89 implementation.

How It Works (n=4 example)

i	a[i]	n-1-i	b[n-1-i]	c[i]
0	1	3	8	9
1	2	2	7	9
2	3	1	6	9
3	4	0	5	9

a = [1,2,3,4], b = [5,6,7,8] → c = [9,9,9,9] ✓

C Program: c[i] = a[i] + b[n-1-i]

/* Given arrays a[] and b[] of n elements, compute c[i] = a[i] + b[n-1-i]
 * Compile: gcc -ansi -Wall -Wextra arrayc.c -o arrayc */
#include <stdio.h>
#define MAX 20

int main(void)
{
    int a[MAX], b[MAX], c[MAX];
    int n, i;

    printf("How many elements (1-%d)? ", MAX);
    if (scanf("%d", &n) != 1 || n < 1 || n > MAX) {
        printf("Enter a number between 1 and %d.\n", MAX);
        return 1;
    }

    printf("Enter %d elements for array a:\n", n);
    for (i = 0; i < n; i++) {
        if (scanf("%d", &a[i]) != 1) { printf("Invalid.\n"); return 1; }
    }

    printf("Enter %d elements for array b:\n", n);
    for (i = 0; i < n; i++) {
        if (scanf("%d", &b[i]) != 1) { printf("Invalid.\n"); return 1; }
    }

    for (i = 0; i < n; i++)
        c[i] = a[i] + b[n - 1 - i];

    printf("\ni    a[i]  b[n-1-i]  c[i]\n");
    printf("---  ----  --------  ----\n");
    for (i = 0; i < n; i++)
        printf("%3d  %4d  %8d  %4d\n", i, a[i], b[n - 1 - i], c[i]);

    return 0;
}

How to Compile and Run

gcc -ansi -Wall -Wextra arrayc.c -o arrayc
./arrayc

Sample Output

How many elements (1-20)? 4
Enter 4 elements for array a:
1 2 3 4
Enter 4 elements for array b:
5 6 7 8

i    a[i]  b[n-1-i]  c[i]
---  ----  --------  ----
  0     1         8     9
  1     2         7     9
  2     3         6     9
  3     4         5     9

How many elements (1-20)? 3
Enter 3 elements for array a:
10 20 30
Enter 3 elements for array b:
1 2 3

i    a[i]  b[n-1-i]  c[i]
---  ----  --------  ----
  0    10         3    13
  1    20         2    22
  2    30         1    31

Code Explanation

Index formula n-1-i — when i=0, n-1-i = n-1 (the last element of b). When i=n-1, n-1-i = 0 (the first element). This maps forward iteration through a[] to reverse iteration through b[]. No reversal of b is needed — the formula handles it implicitly.
Three separate arrays — a[], b[], c[] are all declared. Writing the result into c[] keeps a[] and b[] intact, which allows the table to print both b[n-1-i] and c[i] in the same row. If you modified b[] in place it would corrupt later outputs.
Tabular output — printing i, a[i], b[n-1-i], and c[i] side by side makes it easy to verify each computation. The printf format %3d, %4d, %8d right-aligns the columns for readability.

What This Program Teaches

Index arithmetic — n-1-i is the pattern for “mirror index” — element i from the front corresponds to element n-1-i from the back. The same formula appears in array reversal, palindrome checking, and matrix transposition.
Zip-with-reversed — pairing each element of one sequence with the corresponding element of another sequence (possibly reversed) is a fundamental functional programming concept. In C, it is expressed as a loop with two index expressions.
Preserving input arrays — writing results to a third array is cleaner than modifying the input. It follows the principle of non-destructive computation: the caller can inspect both inputs and the output after the function returns.

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