Question

I was surprised when the following worked

template<typename T>
void f(T &...);

I thought that I have to declare “T” as “typename …T” then, and that it only works in C++0x. But the above compiled in strict C++03 mode. What’s going on?

Answer 1

It’s just the bad old C varargs syntax; the grammar allows omitting the comma. The following are equivalent:

int printf(const char* fmt, ...);
int printf(const char* fmt...);