Does the C ++ standard library have a template getter for the printf conversion specifier?

Question

As we all know, if for some reason you are printf()something, not a stream (which is rare, but can sometimes happen), you need to specify the appropriate format specifier ( dfor signed ints, ufor unsigned, etc., etc.) d.). Now, since it printf()is part of the C ++ standard library, and not just some C legacy, I hope it could be something like

template <typename T> 
const std::string format_specifier;

That will allow, say:

template <typename Foo> 
void bar(const Foo& my_foo) {
    printf(format_specifier<Foo>, my_foo);
}

Is there something similar in the standard library?

Notes:

• , std::cout, , . , ; , : " , ".

Answer 1

, .

- "", ++ .

, C ( , , std ..), , ++ iostream , - .

Answer 2

++ .

-, , ++ , ++ stream. - , , ostream& operator << (ostream&, const T&).

, printf (Python, Java), , , , io-. , :

printf("Operation %-*s : %10.2f$ %2d%/02d\n", size, lib, val, dat->mday, dat->mon + 1);

( (#))

- C ++ - , .

Answer 3

, , :

template<typename T>
char const *format_specifier(); 

template<> inline char const *format_specifier<int>() { return "%d"; }

// ... in function
int d = 5;
printf(format_specifier<decltype(d)>, d); 

IMHO, . , .

The best solution using printf formatting would be code:

my_printf("bla % bla % bla", arg1, arg2, arg3);

where %just gets the correct type from the argument. You can do this using variable templates, and there are already some popular implementations.