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I'm learning c++ and using C++ Primer. Consider the following exercise 14.46:

 class Complex {
     Complex(double);
     // ...
 };

 class LongDouble {

     friend LongDouble operator+(LongDouble&, int);  // (1)

 public:
     LongDouble(int);

     operator double();

     LongDouble operator+(const Complex &);  // (2)
     // ...
  };

 LongDouble operator+(const LongDouble &, double);  // (3)

 LongDouble ld(16.08);
 double res = ld + 15.05; // which operator+ ?

When I compile using gcc 4.5 the above program, I get

14_46.cpp:60:21: error: ambiguous overload for ‘operator+’ in ‘ld + 1.5050000000000000710542735760100185871124267578125e+1’
14_46.cpp:60:21: note: candidates are: operator+(double, double) <built-in>
14_46.cpp:35:5: note:                 LongDouble LongDouble::operator+(const Complex&)
14_46.cpp:45:1: note:                 LongDouble operator+(const LongDouble&, double)
14_46.cpp:17:5: note:                 LongDouble operator+(LongDouble&, int)

Why is (3) not selected ? Isn't it exact match?

However, I noticed that removing const-ness of parameter in (3) matches exactly, i.e.,

LongDouble operator+(LongDouble &, double);  // (4)

Using (4) there is no ambiguity. Am I missing something here?

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2 回答 2

8

您有以下相互竞争的用户定义函数（候选）

operator+(LongDouble&, int); // friend
operator+(LongDouble&, Complex const&); // member
operator+(LongDouble const&, double); // global

您正在使用参数调用它：

(LongDouble&, double)

对于第一个论点，前两个候选人比最后一个更好。对于第二个参数，最后一个候选人比前两个候选人好。对于所有论点，没有候选人至少与所有其他候选人一样好匹配，并且对某些论点有更好的匹配。

对于这种情况，您无法确定一个明显的赢家。这就是我喜欢称之为“纵横交错”的东西。

在用户定义的候选者中也考虑了内置候选者。

operator+(double, double);
operator+(int, double);
...

从所有内置候选人中，operator+(double, double)将最匹配。但这需要对第一个参数进行用户定义的转换，这比第一个参数的所有其他三个用户定义的运算符都差，因此它也无法获胜。

于 2011-07-18T17:04:17.090 回答
2

My preferred way to fix this would be to add const to version 1:

friend LongDouble operator+(const LongDouble&, int);  // (1)
于 2011-07-18T16:59:10.820 回答