Trailing Return Types

[skadush]

Im struggling to understand Trailing return types. Ive bee searching for simple example to understand it but cant find anything


Basically..If i have a function declaration like this
func(int i) -> int (*)[10];

How am i supposed to define it? Is it like this?

auto func(int i) -> int(*)[10]{

   return ?// what is it that im supposed to return?
}

but the problem is I dont know what I am supposed to return? Returning i is invalid.


Again as from my previous questions this is all about the book that i am reading to learn c++ and sometimes doesnt give much information about a certain like this trailing return type that leaves me scratching my head.


This btw about a function that return an array which is illegal in c++ but can attain the result in different way for example

typedef int arrT[10]; 
using arrtT = int[10]; 
arrT* func(int i);

again the author of that book explains what should happen but didnt use it to a usefull example.




So can someone guide me here? Thanks

[MiiNiPaa]

know what I am supposed to return

pointer to array of 10 ints: http://puu.sh/9iHNv/fe080659a3.png
You can manage to return it by returning pointer to static or global array:

auto func(int i) -> int(*)[10]
{
   static int x[10];
   return &x;
}

Your second example declares function returning the same thing, but uses type aliasing to not confuse C++ parser.

[skadush]

@MiiNiPaa

Isnt it illegal to return a reference of local variable of a function?
It would be undefined.

[MiiNiPaa]

It is not local. It is static. It will be created when function first executed and will exist until program finishes. It will not be recreated between function calls and it is a possibility for function to save its state between calls.

[skadush]

I see.. Now i get it.

Thank you very much MiiNiPaa. Helps me alot.