Reverse the elements in an array - c

Question

I've been trying to write a void function called printReverse that takes a double array named d and an int called n as arguments, where n is the number of elements in d. I want printReverse to print the n array values in reverse order. This is what I have so far, but I'm having difficulty declaring arguments.
#include <stdio.h>
int n
double d[n]
void printReverse(d[], double newArray){
int i;
int j;
for (i=0; j=n-1; i<n; i++; j--;){
newArray[j]=d[i];
}
printf("%lf\n",newArray);
}
Any help in what I'm doing wrong would be very much appreciated.
REVISION: Would this code work
void printReverse(double [] d, int n){
int i;
for(i=n-1;i>=0;i--){
printf("%f\n",d[i]);
}

Answer 1

If you want both print and store the array in reverse order you could do something like this:
#include <stdio.h>
void printReverse(double *d, double *newArray, int n){
int i;
int j;
for (i=0, j=n-1; i<n; i++, j--){
newArray[i]=d[j];
printf("%lf\n",newArray[i]);
}
}
int main(){
int n = 4;
double array[] = {1, 2, 3, 4};
double newArr[n];
printReverse(array,newArr,n);
}

Answer 2

//take pointer to array, and size of array as parameters
void printReverse(double* arr,int size){
for(int i =0;i<size;i++){
printf("%f\n",arr[(size-1)-i]);
//(size-1)-i will compute the proper index of the array to read backwards.
//remeber arrays are 0 based meaning that the first member in the array is index 0
}
}
this should do the trick.
EDIT: or if you are trying to reverse the array into a new array and then print...
void printReverse(double* arr,double* arr2,int size){
for(int i =0;i<size;i++){
arr2[(size-1)-i]=arr[i];
}
for (int i=0; i<size; i++) {
printf("%f\n",arr2[i]);
}
}
you will need to have the arrays in the code that calls this function.
so...
double arr[5] = {1,2,3,4,5};
double arr2[5];
printReverse(arr,arr2, 5);