c++ - 使用花括号初始化器列表时模板参数推导失败

Question

我正在尝试在“perpendicular()”函数中使用模板参数推导:

#include <iostream>

template <typename component = double>
struct offset {
  component x;
  component y;
};

template <typename component>
offset(component x, component y) -> offset<component>;

template <typename component>
offset<component> perpendicular(offset<component> const &o) {
  return offset{o.y, -o.x};
}

template <typename component>
std::ostream &operator<<(std::ostream &s, offset<component> const &o) {
  return s << '(' << o.x << ", " << o.y << ')';
}

int main() {
  std::cout << perpendicular({3.1, 1.2}) << '\n';
  return 0;
}

然而，这并不能编译； Clang(带有 -std='c++17' )说: candidate template ignored: couldn't infer template argument 'component' offset<component> perpendicular(offset<component> const &o) { .

我应该放弃写作吗 perpendicular(offset{1.0, 2.0})或者有没有办法给编译器一个提示？

Answer 1

问题 {/*..*/}是它没有类型，而且大多只能推断为 std::initializer_list<T>或 T[N] .

因此以下将允许所需的语法:

template <typename component>
offset<component> perpendicular(component const (&o)[2]) {
    return offset{o[1], -o[0]};
    // return perpendicular(offset{o[0], o[1]});
}

Demo