• Marc (11/19) Jan 14 2018 I call do:
• Adam D. Ruppe (17/29) Jan 14 2018 That doesn't do what I think you think it does.

Marc <jckj33 gmail.com> writes:

let's assume I have

 class C {
   static string foo() { writeln("got called!"); //.... }
 }

then I want to cache foo at some point:

 import std.algorithm;
 auto v = cache(c.foo);

I call do:

 for(int i = 0; i <10; i++) {
 writeln(v);
 }

then it'll print "got called"  only once, which is what I want 
but something obvious is how do I get the returned value as a 
string?
here's why I'm confused more often than I should: I geeeting to 
D's way to do thing and the auto keyword even in documentation 

struct/class returned is explicity so I just navigate to 
aggregate type's documentation page.

Adam D. Ruppe <destructionator gmail.com> writes:

On Monday, 15 January 2018 at 02:54:16 UTC, Marc wrote:
 let's assume I have

 class C {
   static string foo() { writeln("got called!"); //.... }
 }

 then I want to cache foo at some point:

 import std.algorithm;
 auto v = cache(c.foo);


That doesn't do what I think you think it does.

http://dpldocs.info/experimental-docs/std.algorithm.iteration.cache.html

Notice it is from the `iteration` module.. the cache function 
caches results of an iteration, not results of a function. It 
actually returns an object that caches the individual characters 
of that string, rather than the string!

The memoize function is closer to what you want:
http://dpldocs.info/experimental-docs/std.functional.memoize.2.html


Though tbh, I think you should just simply do:

string s = c.foo;
// go ahead and just use s now


 here's why I'm confused more often than I should: I geeeting to 
 D's way to do thing and the auto keyword even in documentation 

 struct/class returned is explicity so I just navigate to 
 aggregate type's documentation page.

The tricky thing is a lot of them create a new type based on its 
arguments, so there would be nothing to navigate to - it all 
depends on what you pass it.

Though, you'll notice with memoize, it returns `ReturnType!fun`, 
that is, the same return type fun (which you passed to it) had.