Function template declaration mystery...

[Steven Schveighoffer]

On 2/28/18 12:47 PM, Robert M. Münch wrote:
> Hi, I'm lost reading some code:
> 
> A a;
> 
> auto do(alias f, A)(auto ref A _a){
>      alias fun = unaryFun!f;
>      return ...
>      ...
> }
> 
> How is this alias stuff working? I mean what's the type of f? Is it an 
> anonymous function which then gets checked to be unary? How is it 
> recognized in the code using the function template?

unaryFun is a template that returns a callable item. It could be a 
struct with an opCall, it could be a function template, it could be an 
alias to a real function, it could be a function pointer, delegate, etc.

It also supports string lambdas, which existed before our current lambda 
syntax.

i.e. alias f = unaryFun!"a - 5";
assert(f(10) == 5);

> 
> 
> This function can be called with code like this:
> 
> a.do((myType) {...myCode...});
> do(a, (myType) {...myCode...});
> 
> What's wondering me here is that the template function only has one 
> paraemter (_a) but I somehow can get my myCode into it. But the code 
> looks like a parameter to me. So why isn't it like:
> 
> auto do(alias f, A)(auto ref A _a, ??? myCode){...
> 
> I'm a bit confused.

This question is a little harder to understand. Perhaps you have real 
code that shows what you are confused about?

-Steve