Array doesn't print in reverse in C using pointers - c

Question

I am trying to make a program that takes 10 numbers as input and outputs them in reverse order using pointers in C.
#include<stdio.h>
#define N 10
int array[N]; //Global variable
int main(void) {
int j;
int i;
printf("Enter 10 numbers: ");
for (i=0;i<N;i++) {
scanf("%d",(array+(4*i))); //Works
}
for (j=N-1;j<0;j--) {
printf("%d",array[j]); //Doesn't print, using *(array+j*4) doesn't
//print also
}
printf("\n");
printf("%d\n",*(array)); //Works so scanf works
printf("%d\n",*(array+4)); //Works so scanf works
return 0;
}
I have tried a making a seperate function for the two for loops but still it doesn't work. I want to know WHY this for-loop doesn't print but the two printfs below it print.
EDIT:
My new code is
#include<stdio.h>
#define N 10
int array[N]; //Global variable
int main(void) {
int j;
int i;
printf("Enter 10 numbers: ");
for (i=0;i<N;i++) {
scanf("%d",(array+i)); //Works
}
for (j=N-1;j<0;j--) { //it is supposed to be j>=0 or j>0 WHY
printf("%d",array[j]); //Doesn't print, using *(array+j) doesn't
//print also
}
printf("\n");
printf("%d\n",*(array)); //Works so scanf works
printf("%d\n",*(array+1)); //Works so scanf works
return 0;
}
Thanks to all the posts, I have a better understanding of how indexing works in C now but the printf doesn't work still unless I change the for-loop conditions(see above). WHY doesn't it work with the initial conditions but with the latter conditions.

Answer 1

Whoa!
This:
scanf("%d",(array+(4*i))); //Works
is very wrong and is overwriting memory! Why are you multiplying the index? You don't need to do that, C can index by itself. It should just be:
scanf("%d", &array[i]);
You want the address of the i:th array member, so say that, don't beat around the bush with strange multiplications.
If you really want to be "using pointers", as mentioned in a comment, you can do so:
scanf("%d", array + i);
This works since array is a pointer to the first element of the array, and adding i to is a fully valid use of pointer arithmetic; C will compute the proper pointer, knowing the size of each int in the array.

Answer 2

Your array consists of 10 elements with type int (obviously). In expression array + i variable i is not an offset in bytes. It is an index of element. So when you read it like you do (scanf("%d",(array+(4*i)))) you basicly read array[0], array[4], array[8], array[12] (we're out of array bounds here already, it causes memory corruption and might cause crashes), etc. Elements array[1],[2],[3],[5], etc. are uninitialized. That's why your code doesn't work :)
UPDATE:
And #shilong-liu's note about array indices is important, too. I didn't notice it.

Answer 3

for (j=N-1;j<0;j--) {
printf("%d",array[j]); //Doesn't print, using *(array+j*4)
}
the for loop is not right. The correct one is that
for (j = N - 1; j > 0; j--)

Answer 4

I guess since the pointer used is of type int, you assume that you have to multiply i by 4 because depending on the compiler int is 4 bytes. I guess if you really care only about the output, then you could do it the way you did with reverse iteration.
What you have to do has been already mentioned by the others so I will give you my solution for actually swapping the pointers memory wise and you could choose from the given solutions:
#include<stdio.h>
#define N 10
int array[N]; //Global variable
int main(void) {
int j;
int i;
printf("Enter 10 numbers: ");
for (i=0; i<N; i++) {
scanf("%d", (array + i));
}
for (left = 0; left < N / 2; left++)
{
int right = N - left - 1;
int temporary = array[left];
array[left] = array[right];
array[right] = temporary;
}
for (i=0; i<N; i++) {
printf("%d", (array + i));
}
return 0;
}
I have been programming for algorithmic contests so you could trust me.