Meaning of ...interface{} (dot dot dot interface)

Below is a piece of Go code I have question about.
Specifically, what is a in this function?

func DPrintf(format string, a ...interface{}) (n int, err error) {
  if Debug > 0 {
    n, err = fmt.Printf(format, a...)
  }
  return
}

Could anyone tell me what the three dots are here?
And what does ...interface{} do?

A parameter type prefixed with three dots (...) is called a variadic parameter. That means you can pass any number or arguments into that parameter (just like with fmt.Printf()). The function will receive the list of arguments for the parameter as a slice of the type declared for the parameter ([]interface{} in your case). The Go Specification states:

The final parameter in a function signature may have a type prefixed with .... A function with such a parameter is called variadic and may be invoked with zero or more arguments for that parameter.

A parameter:

a ...interface{}

Is, for the function equivalent to:

a []interface{}

The difference is how you pass the arguments to such a function. It is done either by giving each element of the slice separately, or as a single slice, in which case you will have to suffix the slice-value with the three dots. The following examples will result in the same call:

fmt.Println("First", "Second", "Third")

Will do the same as:

s := []interface{}{"First", "Second", "Third"}
fmt.Println(s...)

This is explained quite well in the Go Specification as well:

Given the function and calls

func Greeting(prefix string, who ...string)
   Greeting("nobody")
   Greeting("hello:", "Joe", "Anna", "Eileen")

within Greeting, who will have the value nil in the first call, and []string{"Joe", "Anna", "Eileen"} in the second.

If the final argument is assignable to a slice type []T, it may be passed unchanged as the value for a ...T parameter if the argument is followed by .... In this case no new slice is created.

Given the slice s and call

s := []string{"James", "Jasmine"}
   Greeting("goodbye:", s...)

within Greeting, who will have the same value as s with the same underlying array.

Stack Overflow Q#23669720, score: 203