The order of the numbers at the end do not matter, as long as all the even come first.I call this indented code.

Original: 9 8 7 6 5 4 3 2 1Code:#include <stdio.h> #include <stdlib.h> #define N 9 /* MAIN */ void revArray(int arr[]) { int temp[N]; int i; for (i = 0; i < N; ++i) { temp[i] = arr[N - 1 - i]; } for (i = 0; i < N; ++i) { arr[i] = temp[i]; } } int *revElement(int arr[]) { static int idx = N; idx = --idx; return arr + idx; } void evenOdd(int odd[]) { int i; int evenTemp[N / 2]; int oddTemp[(N - 1) / 2]; char tempOdd; for (i = 0; i < N - 1; i++) { if (odd[i] % 2 == 0) /* if even integer */ evenTemp[i] = odd[i]; /*store even numbers temporary */ else oddTemp[i] = odd[i]; /*store odd numbers temporary */ } for (i = 0; i < (N / 2); ++i) { odd[i] = evenTemp[i]; /*swap even numbers first in the array */ odd[i - N] = oddTemp[i]; /*swap odd numbers last in the array */ } } int main(int argc, char **argv) { int a[N]; int idx; int *p = a; while (p < a + N) *p++ = a + N - p; printf("Original: "); p = a; while (p < a + N) printf("%2d ", *p++); /* Part A: Reverse the array */ revArray(a); printf("\nReversed: "); p = a; while (p < a + N) printf("%2d ", *p++); printf("\n"); /* Part B: Return elements in reverse order */ printf("Original: "); for (idx = 0; idx < N; idx++) { printf("%2d ", *revElement(a)); } printf("\n"); /* Part C: Put even numbers first, odd numbers last in the array. Order of the elements is not important as long as all evens are before first odd */ printf("Even: "); evenOdd(a); p = a; while (p < a + N) printf("%2d ", *p++); printf("\n"); system("pause"); }

Reversed: 1 2 3 4 5 6 7 8 9

Original: 9 8 7 6 5 4 3 2 1

Even: 2 4 6 8 5 3 7 9 1

What is the order of the odd numbers at the end of the array? It seems entirely random to me.

Code://!! you're assuming an equal balance of even and odd numbers. //!! what if the input array was all even numbers? int evenTemp[N / 2]; int oddTemp[(N - 1) / 2]; for (i = 0; i < N - 1; i++) { if (odd[i] % 2 == 0) /* if even integer */ //!! you need separate subscripts for each array. //!! at the moment, you make two copies of the input array (which overflow both local arrays) evenTemp[i] = odd[i]; /*store even numbers temporary */ else oddTemp[i] = odd[i]; /*store odd numbers temporary */ }