[Effective C + +--024] If all parameters require type conversion, use the Non-member function for this

Introduction

Let's say we have a class like this:

1 classa{2  Public:3Aintnum =0,intDen =1) {};4     intNum ()Const;5     intDen ()Const;6     ConstAoperator* (Consta& RHS)Const;7};

When doing multiplication, we can take the following actions:

1     A A0 (12); 2     A A1 (13); 3 4     A Match = a0 * A1;     // same-type multiplication 5     Match = match * A0;    // same-type multiplication

The above operation is completely wood-problematic, so what if we want to achieve cross-type multiplication?

So we started trying to manipulate it.

1     2;       // OK        match = a0.operator * 22     2 * A0;       // Error     match = 2.operator * A0

Yes, A0 is a class object that includes the operator* function, so the compiler calls this function. However, 2 does not have a corresponding class, there is no corresponding operator* function.

In the successful operation above, the implicit conversion function is actually called, and in the compiler's opinion, the actual operation is:

1 Const A Temp (2);     // create a temporary object based on 2 2 match = a0 * TEMP;   // equivalent to A0.operator * (temp);

If we replace the constructor of a with the explicit function, then none of the above can be implicitly converted, and no single statement will be compiled.

Let's dig deeper, why match = 2 * A0; can't be done by implicit conversion?

The answer is: This parameter is a qualified participant in an implicit conversion only if the parameter is listed in the parameter column. Status equals "The object that the called member function belongs to"-that is, the metaphor parameter of this object, not a qualified participant. This is why the first compilation passes, and the second compilation does not pass: The first call is accompanied by a parameter placed in the parameter column, and the second call is No.

[Effective C + +--024] If all parameters require type conversion, use the Non-member function for this