C++, what does this syntax mean?

Question

What does this syntax means?

struct actor ... { ... template <typename T0, typename T1> typename result<actor(T0&,T1&)>::type // this line

I know what typename and templates are, my question is about actor(T0&,T1&) syntax

thank you

Blair Holloway · Accepted Answer · 2010-06-09 05:52:20Z

The full declaration from that file reads:

template <typename T0, typename T1> typename result<actor(T0&,T1&)>::type operator()(T0& _0, T1& _1) const { /* snip */ }

If you break it down into chunks, it's easier to read:

template <typename T0, typename T1>

It's a template...

operator()(...) const

...a templated function-call operator function...

operator()(T0& _0, T1& _1) const

...which takes two arguments, by reference, of type T0 and T1...

typename result<...>::type

...and returns a result whose type is the type member of result...

typename result<actor(T0&,T1&)>::type

...where the type of result is paramaterised by the expression actor(T0&,T1&).

R Samuel Klatchko · Accepted Answer · 2010-06-09 05:52:00Z

1

So this means that there is a template called result and within result is a type called type.

template <class T> class result { public: typedef ... type; };

So that line is using that type from the template.

Because the compiler does not know what result<actor(T0&,T1&)>::type is, you need to use typename to tell the compiler to treat it as a type.

Update

actor(T0&,T1&) is a function taking a T0& and a T1& and returning an actor by value.

answered Jun 9, 2010 at 5:44

77k17 gold badges139 silver badges189 bronze badges

4 Comments

what does this mean actor(T0&,T1&)?

@aaa - a function taking a T0 and T1 both by reference and returning an actor by value.

thank you. I am actually somewhat familiar with function/function pointer syntax, but I have never seen it as template parameter.

@aaa to clarify, it's not somehow a function call (as it looks quite similar), but it's the type of a function.