In a C function declaration, what does “…” as the last parameter do?

Question

Often I see a function declared like this:

void Feeder(char *buff, ...)

what does \"...\" mean?

Answer 1

it allows a variable number of arguments of unspecified type (like printf does).

you have to access them with va_start, va_arg and va_end

see http://publications.gbdirect.co.uk/c_book/chapter9/stdarg.html for more information