operator[] 调用者的站点源位置当前解决方法

Question

遗憾的是，当前的源位置不能直接在 operator[] 的参数列表中使用，因为该运算符必须只有一个参数。但是，是否有解决方法，以便我可以获取调用者源代码行？以这段代码为例：

#include <iostream>
#include <string>
#include <source_location>


struct Test
{
  std::source_location src_clone(std::source_location a = std::source_location::current())
  {
    return a;
  }
    
  //this doesnt work:
  //auto operator[](int a, std::source_location src_clone = std::source_location::current())
  auto operator[](int a)
  {
    return std::source_location::current();
  }
};

int main()
{
  auto t = Test{};
    
  auto s1 = t.src_clone();
  std::cout << s1.line() << ' ' << s1.function_name() << '\n';
    
  // is there a way to make this print "main.cpp:30"?
  auto s0 = t[5];
  std::cout << s0.line() << ' ' << s0.function_name() << '\n';
}

Answer 1

找到解决办法：

#include <iostream>
#include <string>
#include <source_location>
#include <string_view>
#include <concepts>

struct string_like
{
  std::string_view strView;
  std::source_location s;
    
  template <typename T>
  string_like (T strView, std::source_location s = std::source_location::current())
    requires std::constructible_from<std::string_view, T>
    : strView(strView), s(s) {}
};

struct Test
{
  auto operator[](string_like s)
  {
    return s.s;
  }
};

int main()
{
  auto t = Test {};
  auto s0 = t["hello"];
 
  // prints main.cpp:29 as it should
  std::cout << s0.line() << ' ' << s0.function_name() << '\n';
}