How to overload a function taking variable number of arguments - c++

Question

I need to distinguish (overload) between two functions - one takes a single const char* argument and the other takes at least two arguments - a const char* followed by one or more arguments.
i.e. basically:
void f(const char *str);
void f(const char *format, ...)
I want the first version to be called for f("hello") and the second version for f("hello %d", 10). The above overload won't work because the compiler finds f("hello") ambiguous.
So I tried this:
void f(const char *str);
template<typename T>
void f(const char *str, T tt, ...);
This makes the overload resolution work correctly. But I end up with another problem. The second function is supposed to forward the arguments for printf-style usage. So I have something like:
template <typename T>
void f ( const char *format, T tt, ... )
{
(T)tt;
va_list varlist;
va_start(varlist, format);
vprintf(format, varlist);
va_end(varlist);
}
Now the second argument tt is no longer part of the variable argument list and calling va_start() with format does not seem to work.
Is there any way to achieve what I want?

Answer 1

If you use a vararg template you can accomplish what you want:
#include <iostream>
void f(const char* str)
{
std::cout << "single arg: " << str << "\n";
}
template <typename ...T>
void f(const char *format, T... args)
{
printf(format, args...);
}
int main()
{
f("hello");
f("formatted %d", 10);
}

Answer 2

Maybe using templates is not the best idea, since the second error seems more difficult to fix than the first. The overload of the functions is fine, it is more if you try in a separate file you will see that it works.
#include <iostream>
bool f(const char* str) {
return true;
}
bool f(const char* str,int n) {
return false;
}
int main(int argc,char* argv[]) {
std::cout << f("Hi") << std::endl; //Gives 1
std::cout << f("Hi",10) << std::endl; //Gives 0
return 0;
}
You could also try creating a namespace and putting the functions in there, thus making sure that no other files share that function name that can cause that error.

Answer 3

Q: why compile error?
A: becauz scope of "void f(const char *format, ...)" had cover "void f(const char *str);".
so the cute solution is you should better using arg num to distinguish 2 func.
like this:
void func_with_one_arg(const char *str) {
printf("func_with_one_arg");
}
void f(int arg_num, const char *format, ...) {
if (arg_num==1) return func_with_one_arg(format);
printf("func_with_multi_arg\n");
va_list ap;
va_start(ap, format);
va_end(ap);
}
if you want to automately coumpute variable arg_num, use this macro to help computing.
#define VA_LENGTH_(_0, _1, _2, _3, _4, _5, _6, _7, _8, _9, _10, _11, _12, _13, _14, _15, _16, _17, _18, _19, _20, N, ...) N
#define VA_LENGTH(...) VA_LENGTH_(0, ## __VA_ARGS__, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0)
#define ExecVF(Func, ...) Func(VA_LENGTH(__VA_ARGS__), __VA_ARGS__)
now you should use ExecVF like this:
int
main(int argc, char **argv) {
ExecVF(f, "foo", "bar");
ExecVF(f, "foo");
return 0;
}