If a class provides both a type conversion whose conversion target is an arithmetic type and an overloaded operator, it will encounter the ambiguity between the overloaded operator and the built-in operator.
If a is a type, the expression a sym b may be:
//The call form cannot be used to distinguish whether the current call is a member function or a non-member function
a.operatorsym(b); //a has an operatorym member function
operatorym( a,b); //operatorsym is an ordinary function
eg.
class SmallInt{
friend SmallInt operator + (const SmallInt&,const SmallInt&);
public:
SmallInt(int = 0);
operator int() const {
return val;}
private:
std::size_t val;
};
SmallInt s1,s2;
SmallInt s3 = s1 + s2; //使用重载的 operator+
//二义性错误,可以把0转换成 SmallInt，然使用 SmallInt 的 +；
//或者把 s3 转换成 int，然后对于两个 int 执行内置的加法。
int i = s3 + 0;
- The set of candidate functions for operators in expressions includes both member functions and non-member functions.
- When an overloaded operator is used to act on an object of a class type, the candidate function contains the ordinary non-member version and the built-in version of the operator. In addition, if the operand on the left is a class type, the overloaded version of the operator defined in the class is also included in the candidate function.
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