*the counter* to reverse an array using pointers doesn't work - c

Question

#include<stdio.h>
int main() {
int i;
int vector[5]={6,17,28,39,410},*r; //variables declaration
r=(int*)&vector; //pointer declaration
for (i = 0; i<5;i++){ //print the array in using a loop
printf("%d ",vector[i]);
}
printf("\n\n");
for(i=0;i<5;i++){ //print the array in reverse order using a loop
vector[i] = *(r+4-i); //it should be from the last to the first but it prints it
printf("%d ",vector[i]); //differently, see below
}
return 0;}
It should be:
6 17 28 39 410
410 39 28 17 6
but it results in:
6 17 28 39 410
410 39 28 39 410
the last two should be 17 6

Answer 1

You are overwriting the data you're trying to read, before reading it.
Just write out the steps your code is taking manually, and you'll see it.
To do reversal in-place, you must swap values around, to avoid overwriting.
Also note that the name of an array evaluates to a pointer to the first argument, in the proper context. So this:
r=(int*)&vector;
is much better written as just:
r = vector;
You should really avoid casts, and your cast is completely unnecessary.

Answer 2

try this:
#include<stdio.h>
int main() {
int i;
int vector[5]={6,17,28,39,410},*r; //variables declaration
r=(int*)&vector; //pointer declaration
for (i = 0; i<5;i++){ //print the array in using a loop
printf("%d ",vector[i]);
}
printf("\n\n");
for(i=0;i<5;i++){ //print the array in reverse order using a loop
//it should be from the last to the first but it prints it
printf("%d ",*(r+4-i)); //differently, see below
}
return ( 0 );
}

Answer 3

In your code you are changing the first two values. So first two steps Your array
will be like this.
410 39 28 39 410
After that loop continues it will get that replaced value. You can store the replaced value in the another array.
for(i=0,j=4;i<j;i++,j--){
temp=vector[i]; // storing the value in temporary variable
vector[i] = r[j];
r[j]=temp; // assigning the value .
}
for ( i=0; i < 5 ; i ++ )
printf("%d\n",vector[i]);