Re: Returning reference: why this works?

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); //

Re: Returning reference: why this works?

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?

Returning reference: why this works?

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 }