Thread overview
Returning reference: why this works?
March 13
import std.stdio;

struct S { int x; }

ref S func1(ref S i) // i is reference
{
    return i;
}

ref S func2(S i) // i is not reference
{
  return func1(i); // Works! Possibility to return reference to local object i?
  //return i; // Error: returning i escapes a reference to parameter i
}

void main() {
  auto ret = func2(S(2));

  writeln(ret); // "S(2)"
}

March 13
On Wednesday, 13 March 2019 at 20:57:13 UTC, Denis Feklushkin wrote:
> import std.stdio;
>
> struct S { int x; }
>
> ref S func1(ref S i) // i is reference
> {
>     return i;
> }
>
> ref S func2(S i) // i is not reference
> {
>   return func1(i); // Works! Possibility to return reference to local object i?

Indeed, you're invoking UB here. With compiler flag `-dip25` that code no longer compiles.

-Johan



March 13
On Wednesday, 13 March 2019 at 21:04:01 UTC, Johan Engelen wrote:
> On Wednesday, 13 March 2019 at 20:57:13 UTC, Denis Feklushkin wrote:
>> import std.stdio;
>>
>> struct S { int x; }
>>
>> ref S func1(ref S i) // i is reference
>> {
>>     return i;
>> }
>>
>> ref S func2(S i) // i is not reference
>> {
>>   return func1(i); // Works! Possibility to return reference to local object i?
>
> Indeed, you're invoking UB here. With compiler flag `-dip25` that code no longer compiles.
>
> -Johan

Oh, very unexpected! Thank you very much!