我想反转我的数组。 为什么这段代码给出垃圾值？

Question

    #include <iostream>


    using namespace std;

    int main(){

    int n;
    int a[n];
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    for(int i=n;i>=0;i--){
       cout<<a[i]<<" ";
   }
    }

Input:- 4 1 2 3 4 Output 4199008 4 3 2 1

Answer 1

For starters the program has undefined behavior because the variable n is not initialized

int n;

So this declaration

int a[n];

is invalid. Moreover variable length arrays is not a standard C++ feature. Instead use the standard class template std::vector .

Also within this loop

for(int i=n;i>=0;i--) {
   cout<<a[i]<<" ";
}

you are trying to access of non-existent element with the index n .

Also you are not reversing an array. You are trying to output an array in the reverse order.

Pay attention to that there are standard algorithms std::reverse and std::reverse_copy declared in the header <algorithm> .

Here is an example how your program with using your approach could look

#include <iostream>
#include <vector>

int main() 
{
    size_t n = 0;

    std::cout << "Enter the size of an array ";

    std::cin >> n;

    std::vector<int> v( n );

    std::cout << "Enter " << n << " elements: ";

    for ( auto &item : v ) std::cin >> item;

    std::cout << "The array in the reverse order\n";

    for ( size_t i = v.size(); i != 0;  )
    {
        std::cout << v[--i] << ' ';
    }
    std::cout << '\n';

    return 0;
}

The program output might look like

Enter the size of an array 10
Enter 10 elements: 0 1 2 3 4 5 6 7 8 9
The array in the reverse order
9 8 7 6 5 4 3 2 1 0 

If to use standard algorithms then your program can look the following way

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>

int main() 
{
    size_t n = 0;

    std::cout << "Enter the size of an array ";

    std::cin >> n;

    std::vector<int> v( n );

    std::cout << "Enter " << n << " elements: ";

    std::copy_n( std::istream_iterator<int>( std::cin ), n, std::begin( v ) );

    std::cout << "The array in the reverse order\n";

    std::reverse_copy( std::begin( v ), std::end( v ), 
                       std::ostream_iterator<int>( std::cout, " ") );
    std::cout << '\n';

    return 0;
}

The program output might look the same way as shown above

Enter the size of an array 10
Enter 10 elements: 0 1 2 3 4 5 6 7 8 9
The array in the reverse order
9 8 7 6 5 4 3 2 1 0 

Answer 2

a[n] will return the element after the last one. When you iterate in reverse order, start with i=n-1 .

Answer 3

At the beginig of your program there is a mistake:

    int n; // You declare n with no value
    int a[n]; // You use is
    cin>>n; // After you used it you get your value-

Now i can suppose that that was just an error while copying it because you give inputs and outputs

Input:- 4 1 2 3 4 Output 4199008 4 3 2 1

So forgeting about that, you declare an array of size n. Remember that the elemnts of the array will go from 0 to n-1. Now look at your second for loop

    // the first element you acces is n and you stop at 1
    // but the array goes from n-1 to 0
    for(int i=n;i>=0;i--){
       cout<<a[i]<<" ";
    }

So you still get n values as an output but the first element that you access is outside of the array. Thats why you get a garbage value, that is a value that was left there.

A solution will be to change the for loop

    for(int i=n-1;i>=-1;i--){
       cout<<a[i]<<" ";
   }

Answer 4

While reversing the array start the loop from n-1 that is i=n-1 (n is the no of elements in array). And run the loop till i>=0. If you will start loop from n it will read illegal index which is out of range and will give you garbage value.

 for(int i=n-1; i>=0; i++){
      cout<<arr[i]<<" ";}