Passing constexpr const char* to template function? - c++

Question

Im trying to wrap a printf like function while preserving the warning you get for invalid format (MSVC doesn't allow extending this to your own printf functions..).
Example here:
https://godbolt.org/z/na_JOe
#include <string>
#include <utility>
#include <stdio.h>
#pragma warning(error : 4477)
constexpr const char* t = "%s";
template<typename... Args>
void test_tpl(Args&&... args) {
constexpr size_t optimistic_buffer_size = 1024;
static thread_local char optimistic_buffer[optimistic_buffer_size];
int ret_count = _snprintf(optimistic_buffer, optimistic_buffer_size, t, std::forward<Args>(args)...);
}
int main() {
std::string hello;
test_tpl(hello);
return 0;
}
The above code works (fires the warning as it should).
However the issue is to make the format string a parameter while preserving the invalid format warning
The following naive approach is invalid syntax:
template<typename... Args>
void test_tpl(constexpr const char* format, Args&&... args) {
Now this works if the string is constexpr - it feels like this should be easy enough to turn into a parameter (although im aware it may not be) ? but is there some simpel way or trick here I can benefit from ?

Answer 1

If I understand correctly... you want to pass the format as argument mantaining the compiler warning about the wrong format...
The only way I see is pass the format as template parameter.
Something as follows
#include <string>
#include <utility>
#include <cstdio>
constexpr char const t[] = "%s";
template <char const * frmt, typename ... Args>
void test_tpl (Args && ... args)
{
constexpr size_t optimistic_buffer_size = 1024;
static char optimistic_buffer[optimistic_buffer_size];
snprintf(optimistic_buffer, optimistic_buffer_size, frmt,
std::forward<Args>(args)...);
}
int main ()
{
std::string hello;
test_tpl<t>(hello);
}