Re: show all subset of a set

mary8shtr@gmail.com writes:


  with throw-away (code)?

#include <iostream> // ::std::cout
#include <ostream> // <<
#include <cassert> // assert
#include <new> // ::std::nothrow

void print( int const set[], size_t const n, size_t const m )
{ size_t j; int * c = new( ::std::nothrow )int[ n + 3 ];
  if( c != nullptr )
  { assert( m >= n ); for( j = 1; j <= n; ++j )c[ j ]= j - 1;
    c[ n + 1 ]= m; c[ n + 2 ]= 0; loop: goto visit;
    next: for( j = 1; c[ j ]+ 1 == c[ j + 1 ]; ++j )c[ j ]= j - 1;
    if( j > n )goto end; c[ j ]= c[ j ]+ 1; goto loop;
    visit: for( size_t i = 1; i <= n; ++i )::std::cout << set[ c[ i ]];
    ::std::cout << "\n";
    goto next; end: delete[] c; }}

int main()
{ int const set[] ={ 1, 2, 3 }; size_t const m = sizeof set / sizeof 0[ set ];
  for( size_t size = 0; size <= m; ++size )print( set, size, m ); }

  The first time ever I used array new and array delete in a program,
  hope I got it right!