Re: What does the function parameter "..." mean?
On 1 Mrz., 07:05, Ivan <linyongt...@gmail.com> wrote:
template<typename T>
class IsClassT {
private:
typedef char One;
typedef struct { char a[2]; } Two;
template<typename C> static One test(int C::*);
template<typename C> static Two test(...); //Here=
, what
does ... mean?
public:
enum { Yes = sizeof(IsClassT<T>::test<T>(0)) == 1 };
enum { No = !Yes };
};
In code, I don't know clear what ... means? But I guess it as:
1) ... means variable argument. But as I known, variable argument is a
placeholder of arguments which must be placed after the last named
argument in C language. Is this ture in C++? if so, ... doesn't means
variable argument.
What is a "variable argument"?
2) ... means wildcard argument. In function overload, if any function
test can't be selected in overload resolution, then the one with ...
as paramenter will be selected.
My guess 1) or 2) is right? If don't, I need your explanation. Thanks!
"..." is known as ellipsis. It's valid in C as well as in C++. It's
sort of a wild card in terms of parameter type AND count of
parameters. Also, it is usually dangerous to use since the compiler
is not able to type check such things. There's a way to access those
parameters if you know their count and type. If the assumption about
parameter count and types is wrong you're in trouble. (Do you
remember printf which can accept any count & types of parameters?)
In this case it's only used for meta programming which is totally
fine. Overload resolution prefers non-ellipsis functions. But in
some cases the ellipsis version will be selected simply because the
first function returning "One" might be ill-formed (for example
there's no type "int double::*").
Cheers!
SG