2

我正在尝试使用spdlog. 我将它与我的代码合并,但现在我收到以下错误:

....fmt\core.h(1016): error C2338: Cannot format argument. To make type T formattable provide a formatter<T> specialization: https://fmt.dev/latest/api.html#formatting-user-defined-types
  ....fmt/core.h(1013): note: while compiling class template member function 'int fmt::v6::internal::arg_mapper<Context>::map(...)'
          with
          [
              Context=context
          ]
  ....fmt/core.h(1213): note: see reference to function template instantiation 'int fmt::v6::internal::arg_mapper<Context>::map(...)' being compiled
          with
          [
              Context=context
          ]
  ....fmt/core.h(1027): note: see reference to class template instantiation 'fmt::v6::internal::arg_mapper<Context>' being compiled
          with
          [
              Context=context
          ]
  ....fmt/core.h(1198): note: see reference to alias template instantiation 'fmt::v6::internal::mapped_type_constant<int,context>' being compiled
  ....fmt/core.h(1342): note: see reference to function template instantiation 'unsigned __int64 fmt::v6::internal::encode_types<Context,int,>(void)' being compiled
          with
          [
              Context=context
          ]
  ....fmt/format.h(3375): note: see reference to class template instantiation 'fmt::v6::format_arg_store<context,int>' being compiled
  ....spdlog/details/fmt_helper.h(49): note: see reference to function template instantiation 'std::back_insert_iterator<fmt::v6::internal::buffer<char>> fmt::v6::format_to<char[6],int&,250,char>(fmt::v6::basic_memory_buffer<char,250,std::allocator<char>> &,const S (&),int &)' being compiled
          with
          [
              S=char [6]
          ]

...这是错误消息的结尾。它永远不会到达我的代码,所以我不知道在哪里看。任何想法可能是什么原因?

spdlog 是 1.6.1 版。最后一个错误行来自这里:

inline void pad2(int n, memory_buf_t &dest)
{
    if (n >= 0 && n < 100) // 0-99
    {
        dest.push_back(static_cast<char>('0' + n / 10));
        dest.push_back(static_cast<char>('0' + n % 10));
    }
    else // unlikely, but just in case, let fmt deal with it
    {
        fmt::format_to(dest, "{:02}", n);  // <--- HERE
    }
}

在我看来并没有什么特别的错误。


更新:在注释掉所有spdlog调用的一些尝试和错误之后,我将其缩小到:

spdlog::info("Foo{}", Point{1, 2});

Point我自己的命名空间中我自己的类在哪里。我确实为它提供了一种打印自己的方法:

template<typename OStream>
OStream &operator<<(OStream &os, const Point &p) {
    return os << "[" << p.x << ", " << p.y << "]";
}
4

1 回答 1

3

spdlog使用该{fmt}库来格式化输出文本,这需要包含<fmt/ostream.h>头文件才能启用std::ostream-like 支持。

无论使用{fmt}库的外部实现,还是 spdlog 源本身的实现,都有<spdlog/fmt/ostr.h>包含正在使用的文件版本的特殊标头。

包含后,spdlog 应该能够使用您的operator<<.

或者,您可以创建一个自定义格式化程序,它也能够解析格式化字符串:

template <>
struct fmt::formatter<Point> {
    constexpr auto parse(format_parse_context& ctx) {
        return ctx.end();
    }

    template <typename Context>
    auto format(const Point& p, Context& ctx) {
        return format_to(ctx.out(), "[{}, {}]", p.x, p.y);
    }
};
于 2020-06-11T14:22:55.327 回答