What's the meaning of `struct decay<T, R(A..., ...)>`

Question

template <typename T, typename R, typename ...A>
struct decay<T, R(A..., ...)> { using type = R(*)(A..., ...); };

What's the exact meaning of that? I need some help~

Answer 1

int foo(int);
int bar(int, ...);

these are two different functions. foo is of type int(int). bar is of type int(int,...).

... is C-style varargs, not to be confused with variardic template arguments which also uses ....

template <typename T, typename R, typename ...A>
struct decay<T, R(A..., ...)> { using type = R(*)(A..., ...); };

This part of an implementation of an optimized version of std::decay within boost::hana. The typename T and T parts are red herrings, part of that optimization.

It is a specialization that matches R(A..., ...), where A... and R are deduced from a function signature.

If you passed double(int, char, ...) as the 2nd argument to this hana::details::decay, R would be double and A... would be int, char. And the ... would "match the C-style varags".

This particular specialization's purpose to to map function signatures that end in C-style varargs to pointers to the same signature. So it maps double(int, char, ...) to double(*)(int, char, ...).

C style varargs are not the same as template variardic arguments. They predate it.

Answer 2

This specialization is one of the specializations that enact the decay of a function type to the corresponding pointer-to-function type, which mirrors the way function lvalues decay to function pointer prvalues.

This particular specialization is used for variable-argument functions (those whose parameter list ends in an ellipsis so that it accepts arguments that don't match any parameters).