如何防止隐式转换double - > int？ [英] How do I prevent an implicit cast double -> int?

问题描述

问题如上，更多细节如下：

Question as above, more details below:

我有一个类 Money 来处理...好了，你猜到了什么。我非常严格不允许 Money 和 double 进行交互（*），因此以下代码

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:

现在我正在考虑允许 Money 与 int ，如你有6块蛋糕，每个$ 4.50（所以去去找到更便宜的蛋糕某处） / p>

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; }

这很好，但 ...
，C ++会从 double 到 int 隐藏转换 ，所以突然我的第一个代码片段编译。我不想要。在这种情况下有没有办法防止隐式强制转换？

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?

谢谢！ - Robin

Thanks! -- Robin

（*）原因：我有很多遗留代码处理所有 Money double ，我不希望这些类型混淆，直到所有运行 Money 。

(*) 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.

编辑：为Money添加构造函数。

Added constructors for Money.

编辑：感谢大家，为您的答案。几乎所有的人都是伟大的和有益的。 R. Martinho Fernandes的评论你可以做 inline const Money operator *（const Money& m，double d）= delete; 其实是答案我切换到支持C ++ 11的编译器）。 Kerrek SB给了一个很好的无C ++ 11的替代品，但是我最终使用的是Nicola Musatti的overload long 方法。这就是为什么我把他的答案标记为答案（也因为所有有用的想法出现作为评论他的答案）。

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!

推荐答案

您可以为扩充赋值运算符的私有重载添加声明：

You could add a declaration for a private overload of your augmented assignment operator:

private:
  Money & operator*=( double i );

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