January 15, 2018
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 confused me a bit as I'm used to C++/C# world where the struct/class returned is explicity so I just navigate to aggregate type's documentation page.

January 15, 2018
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 confused me a bit as I'm used to C++/C# world where the 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.