反向C ++数组

Question

my aim is to reverse an array 3,12,2,1 to 1,2,12,3. when i run this code i get garbage before my actually result. i can't seem to see where the problem is please assit

#include<iostream>
using namespace std;


int rev (int arr[], int a){
    //int r;
    for(int i =a-1; i>=0; i--){
    cout<<arr[i]<<" "; 
    }   
   return 0;
 }

 int main(){
 int arr[] = {6,41,12,5,2};

cout<<"The rev of {6,41,12,5,2}"<<endl;
 cout<<rev(arr, sizeof(arr))<<endl;


   system("pause");
   return 0;
}

Answer 1

An optimized answer to the question would be using reverse () from STL if you are allowed to use it:

std::reverse 

http://www.sgi.com/tech/stl/reverse.html

int main()
{
    int arr[] = {6,41,12,5,2};
    cout<<"The rev of {6,41,12,5,2}"<<endl;
    reverse(arr, arr + 5);
    copy(arr, arr + 5, ostream_iterator<int>(cout, ", "));
}

Answer 2

Use sizeof(arr)/sizeof(arr[0]) instead of sizeof(arr) .

sizeof(arr) gives the total size of the array. sizeof(arr[0]) is the size of one array element (all elements have the same size). So sizeof(arr)/sizeof(arr[0]) is the number of elements.

Answer 3

sizeof return the size in bytes. In your example, if sizeof(int) = 4, it returns 20.

Answer 4

Because you're using an array, you have to keep the size of the array handy as well. sizeof computes the size of a value in memory, in this case the size of all the memory used to represent arr . You can do sizeof(arr)/sizeof(int) to get the number of elements in an array. This makes sense because it's taking the total size of the array and dividing it by the size of an element in the array. Beware however that this only works for arrays ( int arr[4] = {6,41,12,5,2}; ). If it's a pointer to a heap-allocated array via something like int* i = new int[4]; you'll need to keep the size of the array hanging around.

Also, you're calling your reverse function from within a cout<< call, which will print the function's return value (in this case it's hard-coded to 0).

It also turns out there is a function in the C++ standard library ( std::reverse ) that can do this.

Answer 5

If I may speak subjectively and in an off-topic manner about your approach, it is very un-C-like. My personal favorite way to reverse an array goes like this:

void reverse(int *a, int n)
{
   int *p = a, *q = a + n - 1;

   while (p < q)
   {
      int swap = *p;
      *p++ = *q;
      *q-- = swap;
   }
}

// Usage:

   int a [] = { /* ... */ };

   reverse(a, sizeof(a)/sizeof(*a));

Of course, since your question is tagged c++ , there's always std::reverse() .

Answer 6

sizeof运算符返回一个额外的值（arrayLength + 1），这里6将在传递6时返回，存储在a-1中时，您将得到5，但数组索引从0长度1开始，即从0到4，这里我指向索引5不是最后一个元素last + 1为何获得垃圾值