c++ - 在多参数情况下使用转换构造函数

Question

我在看这个视频
CppCon 2018:丹·萨克斯“结交新 friend ”please look here
在这次谈话中，他指的是(假设的)有理数类(有
分子和分母作为私有(private)成员，所有计算都围绕
围绕这个。我正在尝试实现本次演讲中提到的任何内容
很好地理解模板。
我说，Dan 提到了一个执行乘法的函数 int*rational

//~ rational.h
template <typename> class rational;
template <typename T> std::ostream& operator<<(std::ostream&, const rational<T> &);
template <typename T> rational<T> operator*(const rational<T>&, const rational<T>&);

template <typename T>
class rational {
    friend std::ostream& operator<<<T>(std::ostream&, const rational<T>&);
    friend rational<T> operator*<T>(const rational<T>&, const rational<T>&);
    public:
        rational(); 
        rational(T num);
        rational(T num, T den); 
        rational(const rational &ro);

        rational &operator+=(const rational &ro);
        rational &operator*=(const rational &ro);
...
...
    private:
        T num, den;
};

// implements operator*
template <typename T>
rational<T> operator*(const rational<T> &lo, const rational<T> &ro) {
    rational<T> result(ro);
    result *= lo;
    return result;
}

然后，我们有 main

//~ temp.cpp
#include <iostream>
#include "rational.h"

using namespace std;

int main(int argc, char *argv[]) {
    rational<int> r1(1,3);
    rational<int> r2(5,6);

    cout << "r1=" << r1 << ", r2=" << r2 << endl;

    rational<int> r3(3);

    r3 = 3*r3;

    cout << r3 << endl;

    return 0;
}

但是，当我构建它时，我得到了错误:

onkar@TITAN:~/cc/test$ make
g++ -g -Wno-unused-variable -Wno-unused-but-set-variable -std=c++11 -Wall -c test.cpp
test.cpp: In function ‘int main(int, char**)’:
test.cpp:14:8: error: no match for ‘operator*’ (operand types are ‘int’ and ‘rational<int>’)
   14 |  r3 = 3*r3;
      |       ~^~~
      |       | |
      |       | rational<int>
      |       int
In file included from test.cpp:2:
rational.h:58:13: note: candidate: ‘template<class T> rational<T> operator*(const rational<T>&, const rational<T>&)’
   58 | rational<T> operator*(const rational<T> &lo, const rational<T> &ro) {
      |             ^~~~~~~~
rational.h:58:13: note:   template argument deduction/substitution failed:
test.cpp:14:9: note:   mismatched types ‘const rational<T>’ and ‘int’
   14 |  r3 = 3*r3;
      |         ^~
make: *** [Makefile:2: test] Error 1
onkar@TITAN:~/cc/test$ 

现在，我确实理解了这个错误(转换构造函数采用单个参数和所有参数)
但问题是:
如何确定 r3 = 3*r3;有效(而且我不必打电话(看起来很奇怪) r3 = rational<int>(3)*r3反正我总能做到)。

Answer 1

template argument deduction 中不考虑隐式转换.给定 3*r3 , T 的模板参数推导基于第一个函数参数(即 3 )失败；来自 int 的隐式转换至rational<int>不会考虑。

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

您可以更改 operator*到非模板，它没有这样的问题，隐式转换将按您的预期工作。

template <typename> class rational;
template <typename T> std::ostream& operator<<(std::ostream&, const rational<T> &);

template <typename T>
class rational {
    friend std::ostream& operator<<<T>(std::ostream&, const rational<T>&);
    friend rational<T> operator*(const rational<T> &lo, const rational<T> &ro) {
        rational<T> result(ro);
        result *= lo;
        return result;
    }
    public:
        rational(); 
        rational(T num);
        rational(T num, T den); 
        rational(const rational &ro);

        rational &operator+=(const rational &ro);
        rational &operator*=(const rational &ro);
...
...
    private:
        T num, den;
};