Intro

With the recent release of GCC 13 and Clang 17, <format> is finally avaiable to the most of us. I’m using GCC and got some hand-on experience with it.

I’ve been a long time user of fmtlib, the library which the standard <format> is based on. So it’s natural for me to try to use <format> the way I use <fmt/core.h>; can it be done by simply replacing fmt:: with std::?

Formatting User Type

Printing user types has been a common task. In the past it is done by overloading operator<<. With fmtlib, it is done by providing a formatter class specialization.

The example I’ve been using since 2019 is:

// the type I want to format
struct Point {
    int x, y;
};


namespace fmt {

// formatter for it
template<>
struct formatter<Point> {
    template <class Context>
    auto parse(Context& ctx) { return ctx.begin(); }

    template <class Context>
    auto format(const Point& p, Context& ctx) {
        return format_to(ctx.out(), "Point({}, {})", p.x, p.y);
    }
};

}

Does it work if I simply change namespace fmt to namespace std?

Unfortunately, no. The first error I got is:

C:/msys64/ucrt64/include/c++/13.1.0/format:3595:47: error: call to non-'constexpr' function 'auto std::formatter<Point>::parse(Context&) [with Context = std::basic_format_parse_context<char>]' 3595 |               this->_M_pc.advance_to(__f.parse(this->_M_pc));

It seems quite clear - the compiler wants parse to be constexpr. This makes sense, as it may want to do some compile-time validation. No problem. So I modified my code to be:

    template <class Context>
    constexpr auto parse(Context& ctx) { return ctx.begin(); }
//  ^^^^^^^^^

and compile again.

It still doesn’t work. And there are walls of error texts.

Thankfully, as C++ programmers we’re already used to dealing with walls of error text. Here was my then throught process:

The direct error is:

C:/msys64/ucrt64/include/c++/13.1.0/format:3241:38: error: no matching function for call to 'std::basic_format_arg<std::basic_format_context<std::__format::_Sink_iter<char>, char> >::basic_format_arg(Point&)'

This doesn’t tell me much. So I keep reading.

Somewhere down the line:

C:/msys64/ucrt64/include/c++/13.1.0/format:2996:9: note:   template argument deduction/substitution failed:
C:/msys64/ucrt64/include/c++/13.1.0/format:2996:9: note: constraints not satisfied
...
C:/msys64/ucrt64/include/c++/13.1.0/format:2201:20: note: the required expression '__cf.format(__t, __fc)' is invalid
 2201 |       { __cf.format(__t, __fc) } -> same_as<typename _Context::iterator>;

I assume __cf is my formatter, __t is a Point, and __fc is the formatter context. But how is its return type not iterator though?

Keep reading, and I found

cc1plus.exe: note: set '-fconcepts-diagnostics-depth=' to at least 2 for more detail

So I compile again with -fconcepts-diagnostics-depth=2. This time it includes a new error:

C:/msys64/ucrt64/include/c++/13.1.0/format:2201:20: error: passing 'const std::formatter<Point>' as 'this' argument discards qualifiers [-fpermissive]

And that’s why! The member function format(...) needs to be const. Now everything compiles.

namespace std {

template<>
struct formatter<Point> {
    template <class Context>
    constexpr auto parse(Context& ctx) { return ctx.begin(); }

    template <class Context>
    auto format(const Point& p, Context& ctx) const {
        return format_to(ctx.out(), "Point({}, {})", p.x, p.y);
    }
};

}

To be fair, after this I checked the fmtlib documentation again, and the up-to-date user type formatter example does include constexpr and const. I guess I’ll just have to keep these in mind from now on.

Format String Forwarding

While std::format is here, std::print is not. This means that we have to write

std::cout << std::format("The answer is {}\n", 42);

for now.

Still, it is desirable to provide a poor man’s print to make it convenient:

template<class FmtString, class ...Args>
void print(FmtString fmt, Args&&... args) {
    std::cout << std::format(fmt, std::forward<Args>(args)...);
}

int main() {
    print("This answer is {}\n", 42);
}

Unfortunately, the above code doesn’t compile. The error is:

formatter.cpp: In instantiation of 'void print(FmtString, Args&& ...) [with FmtString = const char*; Args = {int}]':
formatter.cpp:32:10:   required from here
formatter.cpp:26:29:   in 'constexpr' expansion of 'std::basic_format_string<char, int>(fmt)'
formatter.cpp:26:29: error: 'fmt' is not a constant expression
   26 |     std::cout << std::format(fmt, std::forward<Args>(args)...);
      |

This one took a little bit of digging. It turns out std::format only supports compile-time format string, as it is mandated to perform format string checking at compile time.

The way it does this is to have a format string type that depends on the arguments (format_string<Args...>), so it can “remember” the argument types. The constructor of format_string is consteval and it cannot read through an “outside” const char*, which our print essentially does, hence the error.

The make our print work:

template<class ...Args>
void print(std::format_string<Args...> fmt, Args&&... args) {
    std::cout << std::format(fmt, std::forward<Args>(args)...);
}

Afterword

It is great to have std::format available, even though it is quite restricted.

If any of the following is required:

  • C++17 or before
  • Dynamic format string
  • print

Then only fmtlib can fufill the requirements.

For reasons above, I expect fmtlib to continue to be widely used in the near future.