• Dfr (7/7) Dec 25 2013 Hello, following code:
• H. S. Teoh (9/20) Dec 25 2013 joiner returns a range object, not an array. To get an array out of it,
• lomereiter (5/12) Dec 25 2013 std.conv.to should be used instead:

"Dfr" <deflexor yandex.ru> writes:

Hello, following code:

import std.algorithm : joiner;
string joined = joiner(["hello", "world"], " ");

Results in:

Error: cannot implicitly convert expression (joiner(...)) of type 
Result to string

Any idea how to make this work ?

"H. S. Teoh" <hsteoh quickfur.ath.cx> writes:

On Wed, Dec 25, 2013 at 06:21:17PM +0000, Dfr wrote:
 Hello, following code:
 
 import std.algorithm : joiner;
 string joined = joiner(["hello", "world"], " ");
 
 Results in:
 
 Error: cannot implicitly convert expression (joiner(...)) of type
 Result to string
 
 Any idea how to make this work ?

joiner returns a range object, not an array. To get an array out of it,
do this:

	import std.array : array;
	import std.algorithm : joiner;
	string joined = joiner(["hello", "world"], " ").array;


T

-- 
Spaghetti code may be tangly, but lasagna code is just cheesy.

"lomereiter" <lomereiter gmail.com> writes:

On Wednesday, 25 December 2013 at 18:41:47 UTC, H. S. Teoh wrote:
 	import std.array : array;
 	import std.algorithm : joiner;
 	string joined = joiner(["hello", "world"], " ").array;


 T

Ha!
  Error: cannot implicitly convert expression 
 (array(joiner(["hello", "world"], " "))) of type dchar[] to 
 string

std.conv.to should be used instead:

     import std.conv : to;
     string joined = joiner(["hello", "world"], " ").to!string;