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c++ - How do I prevent an implicit cast double -> int?

Question as above, more details below:

I have a class Money to deal with... well, you guessed what. I am very strict about not allowing Money and double to interact(*), so the following code is not possible:

Money m1( 4.50 );
double d = 1.5;
Money m2 = m1 * d; // <-- compiler error

Now I'm thinking about allowing multiplication of Money with int, as in "you have 6 pieces of cake for $4.50 each (so go and find cheaper cake somewhere)."

class Money
{
    Money();
    Money( const Money & other );
    explicit Money( double d );
    ...
    Money & operator*=( int i );
    ...
}
inline const Money operator*( const Money & m, int i ) { return Money( m ) *= i; }
inline const Money operator*( int i, const Money & m ) { return Money( m ) *= i; }

That works fine, but... unfortunately, C++ does implicit casts from double to int, so suddenly my first code snippet will compile. I don't want that. Is there any way to prevent implicit casts in this situation?

Thanks! -- Robin

(*) Reason: I have lot's of legacy code that handles all Money-related stuff with double, and I don't want those types confused until everything run with Money.

Edit: Added constructors for Money.

Edit: Thanks, everyone, for your answers. Almost all of them were great and helpful. R. Martinho Fernandes' comment "you can do inline const Money operator*( const Money & m, double d ) = delete;" was actually the answer (as soon as I switch to a C++11-supporting compiler). Kerrek SB gave a good none-C++11 alternative, but what I ended up with using is actually Nicola Musatti's "overload long" approach. That's why I'm flagging his answer as "the answer" (also because all the useful ideas came up as comments to his answer). Again, thanks!

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How about a template plus compile-time trait check:

#include <type_traits>

// ...

template <typename T>
Money & operator*=(const T & n)
{
  static_assert(std::is_integral<T>::value, "Error: can only multiply money by integral amounts!");
  // ...
}

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