c++ - 如何使用模板函数进行隐式转换

Question

非常简化示例(不管类A和运算符在做什么，这只是示例):

#include <iostream>
using namespace std;

template <bool is_signed>
class A {
public:
    // implicit conversion from int
    A(int a) : a_{is_signed ? -a : a}
    {}

    int a_;
};

bool operator==(A<true> lhs, A<true> rhs) {
    return lhs.a_ == rhs.a_;
}

bool operator==(A<false> lhs, A<false> rhs) {
    return lhs.a_ == rhs.a_;
}

int main() {
    A<true> a1{123};
    A<false> a2{123};

    cout << (a1 == 123) << endl;
    cout << (a2 == 123) << endl;

    return 0;
}

这有效。

但是如果我用模板替换两个 operator==(具有相同的主体):

template <bool is_signed>
bool operator==(A<is_signed> lhs, A<is_signed> rhs) {
    return lhs.a_ == rhs.a_;
}

，其编译产生错误:

prog.cpp: In function ‘int main()’:
prog.cpp:31:14: error: no match for ‘operator==’ (operand types are ‘A<true>’ and ‘int’)
  cout << (a1 == 123) << endl;
           ~~~^~~~~~
prog.cpp:23:6: note: candidate: ‘template<bool is_signed> bool operator==(A<is_signed>, A<is_signed>)’
 bool operator==(A<is_signed> lhs, A<is_signed> rhs) {
      ^~~~~~~~
prog.cpp:23:6: note:   template argument deduction/substitution failed:
prog.cpp:31:17: note:   mismatched types ‘A<is_signed>’ and ‘int’
  cout << (a1 == 123) << endl;
                 ^~~

这里可以使用模板吗？我可以以某种方式使用 C++17 用户定义的模板推导指南吗？还是其他什么？

Answer 1

template argument deduction 中不考虑隐式转换，这会导致扣除 is_signed第二个函数参数失败。

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

如果您始终使用operator==风格如 a1 == 123 ，即 A<is_signed>始终用作第一个操作数，您可以从推导中排除第二个函数参数。例如

template <bool is_signed>
bool operator==(A<is_signed> lhs, std::type_identity_t<A<is_signed>> rhs) {
    return lhs.a_ == rhs.a_;
}

LIVE

_{PS: std::type_identity 从 C++20 开始支持；即使实现起来并不难。}