void main()
{
int a[10],i,n;
printf("enter howmany elements");
scanf("%d",&n);

for(i=0;i<=n;i++)
scanf("%d",&a[i]);

for(i=n;i>=0;i--)
printf("%d",a[i]);

}

The above solution is ritee.. :-)

cheers..

#include<stdio.h>
int main()
{
 int a[]={1,2,3,7,8};
 int count=0,i;
 while(a[i] != '\0')
 {
    count++;
 }
 for(i=count;i>=1;i--)
 { 
    printf("%d",a[i]);
 }
}

Both the solution above is wrong...

Answer #1: It will input an extra element in array coz array is starting from 0 and goes till n so total elements n+1

Answer #2: C Automatically starts array from 0 index and not 1 if not programmed otherwise and in the given program we are starting the loop from 1

Solution

void main()
{
int a[30],n;
scanf("%d",&n);

for(int i=0;i<n;i++)
scanf("%d",&a[i]);

for(i=n-1;i>=0;i--)
printf("%d",a[i]);

}
}

in the above of my answer is correct coz in that i am taking
count =0;
so the array will start from last element.

Answer #2: You have assigned the elements of the array in 
the program itself which by default starts from array index 
0 that is the numbers goes into  the array indices 0 1 2 3 
4 and you have not initialzed or modified the value of i in 
the first loop  which happens to be a garbage or 0 so the 
count goes to infinite. coz a[i] is never '\0'
 Secondly in a numeric array last element in not null 
character it is only in case of character arrays.
Assuming that you have implemented the 1st loop correctly 
and you have the length of array a in count ie = 4 since u 
are starting from 0 then you have initialized i to count 
which makes i = 4.In the second loop the condition is i>=1 
so i=0 is never visited and thus a[0] ir not displayed 
(remember in C array starts from 0 by default).