• Chris Katko (4/4) Jun 18 2022 ````D
• Adam D Ruppe (9/13) Jun 18 2022 An array of pair is `pair[]`, keep the brackets with the type.
• Chris Katko (6/20) Jun 18 2022 Thanks!!

Chris Katko <ckatko gmail.com> writes:

````D
struct pair { float x, y;}

pair p[] = [[0, 0], [255, 255], [25,-25]]; //nope
````

Adam D Ruppe <destructionator gmail.com> writes:

On Saturday, 18 June 2022 at 17:37:44 UTC, Chris Katko wrote:
 ````D
 struct pair { float x, y;}

 pair p[] = [[0, 0], [255, 255], [25,-25]]; //nope
 ````

An array of pair is `pair[]`, keep the brackets with the type.

Then a struct literal is either:

pair(0, 0) // using constructor syntax

or in some select contexts (including this one):

{0, 0} // using named literal syntax


Therefore:

pair[] p = [{0, 0}, {255, 255}, {25,-25}];

compiles here.

Chris Katko <ckatko gmail.com> writes:

On Saturday, 18 June 2022 at 17:52:16 UTC, Adam D Ruppe wrote:
 On Saturday, 18 June 2022 at 17:37:44 UTC, Chris Katko wrote:
 ````D
 struct pair { float x, y;}

 pair p[] = [[0, 0], [255, 255], [25,-25]]; //nope
 ````

 An array of pair is `pair[]`, keep the brackets with the type.

 Then a struct literal is either:

 pair(0, 0) // using constructor syntax

 or in some select contexts (including this one):

 {0, 0} // using named literal syntax


 Therefore:

 pair[] p = [{0, 0}, {255, 255}, {25,-25}];

 compiles here.

Thanks!!

One extra caveat for anyone who googles this. You can't use {0, 
0} notation if the struct has constructors. Which make sense 
since you don't want people accidentally bypassing a constructor 
if it exists.