#include <iostream>
using namespace std;

int* reverse(int arr[],int n){
    int rev[100];
    int j =0;
    for(int i=n-1;i>=0;i--){
            rev[j]=arr[i];
            j  ;
            }
    return rev;
}
int main() {
    int n;
    cin>>n;
    int arr[100];
    for(int i=0;i<n;i  ){
        cin>>arr[i];
    }
    cout<<reverse(arr,n);
}

I am trying reverse an array using loops but don't know what the error was it was returning some bin value.

CodePudding user response：

Your rev temporary resides in automatic storage. It means that the object will be gone after the function returns. While C allows you to decay rev to an int* and then return said pointer, it does not mean that this returns the object itself. You merely get a pointer to an already destroyed object. Not very useful. In fact, doing anything with this pointer will cause undefined behaviour.

Usually what you want to do is reverse things in-place. That's also how std::reverse works.

So, there are two options. If you have a completely filled c-style array, you could write a reverse function like this:

template <std::size_t N>
void reverse(int (&a)[N]) {
  // reverse a from 0 to N-1
}
reverse(a);

Or, if you have an only partially filled array, take a page out of the standard library and reverse a range, denoted by two iterators.

void reverse(int* begin, int* end) {
  /* begin points to the first entry, end points one past the last */
}
reverse(a, a n);

Of course, instead of using c-style arrays, you could use a dynamically growing array such as std::vector, which carries the actual size of the array around for you.