Re: Returning reference: why this works?

[Denis Feklushkin via Digitalmars-d-learn]

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!

Re: Returning reference: why this works?

[Johan Engelen via Digitalmars-d-learn]

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

Returning reference: why this works?

[Denis Feklushkin via Digitalmars-d-learn]

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)"
}