[英]std::format of user-defined types?
In C++20 - how do you make a user-defined type compatible with std::format
?在 C++20 中 - 如何使用户定义的类型与std::format
兼容?
For example, let's say I have a type called Point
:例如,假设我有一个名为Point
的类型:
struct Point {
int x;
int y;
};
with its operator<<
defined:其operator<<
定义:
inline std::ostream&
operator<<(std::ostream& o, Point pt)
{ return o << "[" << pt.x << << ", " << pt.y << "]"; }
then will the following program output Hello [3, 4]!
那么下面的程序会输出Hello [3, 4]!
? ?
int main() {
Point pt{3,4};
std::cout << std::format("Hello {}!\n", pt);
}
If yes - why and how?如果是 - 为什么以及如何?
If no - what do I have to add to the definition of Point
to to make it work?如果不是 - 我必须在Point
to 的定义中添加什么才能使其工作?
std::format
doesn't support operator<<
, you need to provide a formatter
specialization for your type ( Point
) instead. std::format
不支持operator<<
,您需要为您的类型( Point
)提供formatter
专业化。 The easiest way to do it is by reusing one of existing formatters, eg std::formatter<std::string>
:最简单的方法是重用现有的格式化程序之一,例如std::formatter<std::string>
:
template <>
struct std::formatter<Point> : std::formatter<std::string> {
auto format(Point p, format_context& ctx) {
return formatter<string>::format(
std::format("[{}, {}]", p.x, p.y), ctx);
}
};
This will give you all format specifications supported by std::string
out of the box.这将为您提供开箱即用的std::string
支持的所有格式规范。 Here is an example of formatting Point
with center alignment padded with '~' to 10 characters:下面是一个使用“~”填充中心对齐到 10 个字符的格式化Point
示例:
auto s = std::format("{:~^10}", Point{1, 2});
// s == "~~[1, 2]~~"
which is nontrivial to achieve with iostreams.使用 iostreams 实现这一点非常重要。
You have to specialize std::formatter
for your type.您必须为您的类型专门化std::formatter
。
namespace std
{
template<class CharT>
struct formatter<Point, CharT>
{
template <typename FormatParseContext>
auto parse(FormatParseContext& pc)
{
// parse formatter args like padding, precision if you support it
return pc.end(); // returns the iterator to the last parsed character in the format string, in this case we just swallow everything
}
template<typename FormatContext>
auto format(Point p, FormatContext& fc)
{
return std::format_to(fc.out(), "[{}, {}]", p.x, p.y);
}
};
}
I don't think the ostream operator will work but I have no sources to support this claim.我不认为 ostream 操作符会起作用,但我没有支持这种说法的来源。
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