Re: "Swapping" two arrays like this doesn't compile because..

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 22 Feb 2007 06:38:20 -0500

Message-ID:

<erjvbc$rvr$1@murdoch.acc.Virginia.EDU>

Eric Lilja wrote:

..during the call to swap_arrays the name of the array decays to a
pointer to its first element and that pointer is a temporary variable
and you cant bind a non-const reference to a temporary. Correct
analysis?

I don't think so.

The code (that only compiles if you comment out the call to
swap_arrays()):
#include <iostream>
#include <iterator>
using namespace std;

void swap_arrays(int *&a1, int *&a2)
{
   int *ptr = a1;
   a1 = a2;
   a2 = ptr;
}

That function will not swap arrays. It swaps objects of type pointer-to-int.

int main()
{
   int array1[] = {1, 2, 3};
   int array2[] = {4, 5, 6};

   swap_arrays(array1[0], array2[0]);

You promised int*& as the parameter. What you pass is int&.

   cout << "array1 after swap:" << endl;

   copy(array1, array1 + sizeof(array1) / sizeof(array1[0]),
ostream_iterator<int>(cout, " "));

   cout << endl << "array2 after swap:" << endl;

   copy(array2, array2 + sizeof(array2) / sizeof(array2[0]),
ostream_iterator<int>(cout, " "));

   cout << endl;
}

Try this:

#include <cstddef>
#include <algorithm>

using namespace std;

template < typename T, std::size_t N >
void swap ( T (&lhs) [N], T (&rhs) [N] ) {
  for ( std::size_t i = 0; i < N; ++i ) {
    swap( lhs[i], rhs[i] );
  }
}

#include <iostream>
#include <iterator>

using namespace std;

int main ( void ) {
  int array1[] = {1, 2, 3};
  int array2[] = {4, 5, 6};

  swap( array1, array2 );

   cout << "array1 after swap:" << endl;

   copy(array1, array1 + sizeof(array1) / sizeof(array1[0]),
        ostream_iterator<int>(cout, " "));

   cout << endl << "array2 after swap:" << endl;

   copy(array2, array2 + sizeof(array2) / sizeof(array2[0]),
        ostream_iterator<int>(cout, " "));

   cout << endl;
}

Best

Kai-Uwe Bux