Variadic template with printf argument checking

Question

I have a printf-style function that takes a variable number of arguments. Here is my starting point:

#include <stdio.h>
#include <stdarg.h>
void MyPrint (const char* fmt,...)
  {
  va_list arglist ;
  va_start (arglist, fmt) ;
  vprintf (fmt, arglist) ;
  va_end (arglist) ;
  }

int main()
  {
  MyPrint ("Hello, %s\n", "world") ;
  }

This prints Hello, world as expected.

Now I want to make two changes. First, I want to check the format string using the format attribute of g++. So I declare the MyPrint function first (I have to declare it first, because for some reason g++ doesn't let you assign attributes to a function definition):

void MyPrint (const char* fmt,...) __attribute__ ((format (printf, 1, 2))) ;

Now if I try e.g. MyPrint ("Hello, %d\n", "world") ; I get a nice error message.

The second change I want to make is to use a variadic template parameter. Like this:

#include <utility> // for std::forward
template<typename...Params>
void MyPrint (Params&&... fmt)
  {
  printf (std::forward<Params> (fmt)...) ;
  }

This works too. So I combine the two, by adding the format-checking attribute to the variadic function template with this forward declaration:

template<typename...Params>
void MyPrint (Params&&... fmt) __attribute__ ((format (printf, 1, 2))) ;

But now I get this error message (gcc 10.2):

<source>: In substitution of 'template<class ... Params> void MyPrint(Params&& ...) [with Params = {const char (&)[11], const char (&)[6]}]':
<source>:15:38: required from here
<source>:8:6: error: 'format' attribute argument 2 value '1' refers to parameter type 'const char (&)[11]'

This has got me completely baffled. Can anybody tell me what I'm doing wrong?

Here is the complete program:

#include <stdio.h>
#include <utility> // for std::forward

template<typename...Params>
void MyPrint (Params&&... fmt) __attribute__ ((format (printf, 1, 2))) ;

template<typename...Params>
void MyPrint (Params&&... fmt) // <-- Line 8
  {
  printf (std::forward<Params> (fmt)...) ;
  }

int main()
  {
  MyPrint ("Hello, %s\n", "world") ; // <-- Line 15
  }

Does std::forward make any sense here? It's going to be copied (and even promoted) to printf anyway. The second change I want to make is to use a variadic template parameter what for? What's the goal here? — KamilCuk
– KamilCuk, Commented Feb 28, 2021 at 12:38
@KamilCuk: "What's the goal here?" The function is actually a class constructor. I want derived classes to have similar variadic constructors, that can simply call the base class's constructor 'as is'. — TonyK
– TonyK, Commented Feb 28, 2021 at 13:33

user14215102 · Accepted Answer · 2021-02-28 12:35:06Z

2

You can make the first error go away by adding a fixed const char * argument as the format string and pointing the attribute to that.

template<typename...Params>
void MyPrint (const char * format, Params&&... fmt) __attribute__ ((format (printf, 1, 2))) ;

template<typename...Params>
void MyPrint (const char * format, Params&&... fmt) // <-- Line 9
  {
  printf (format, std::forward<Params> (fmt)...) ;
  }

Which reveals another error:

test.cc:8:6: error: ‘format’ attribute argument 3 value ‘2’ does not refer to a variable argument list
    8 | void MyPrint (const char * format, Params&&... fmt) // <-- Line 9
      |      ^~~~~~~

It seems that the attribute for checking the printf archetype relies on one const char * argument and a variable argument list and is not willing to work without them. So you have to give up either the C++ template magic or the compile-time format string checking.

answered Feb 28, 2021 at 12:35

user14215102

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2 Comments

TonyK Over a year ago

Is there a logical reason that this is impossible, do you think? Or is it just a shortcoming of the compiler? (I tried various compilers, but they all reject my code.)

user14215102 Over a year ago

I don't think it's impossible, it's just not supported. If you have a compelling argument why this would be useful, you can try raising it as a wishlist bug against gcc.

TonyK · Accepted Answer · 2023-08-27 10:40:15Z

0

Just for info: this has been fixed in gcc 13.1. The program works as expected, with no compiler errors or warnings.

answered Aug 27, 2023 at 10:40

TonyK

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Comments

Fei Du · Accepted Answer · 2024-09-12 22:41:42Z

0

Do not use printf. Use C++23 std::print() or C++20 std::format()

The following code can be compiled in clang 18 with -std=c++23 -stdlib=libc++.

#include <print>  // C++23
#include <utility>

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

int main() {
    my_print("Hello, {}\n", "world");
}

answered Sep 12, 2024 at 22:41

Fei Du

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