The lazy val is declared as a member of the function, so each time the
function is called there is a new lazy val being memoized.
Brian Maso
On Tue, Apr 10, 2018 at 8:19 AM, <des.mag...@gmail.com> wrote:
> Hi
>
> Based on my understanding of the explanation of lazy val memoization in
> section 5.2.1, I would have thought that if I define the following function:
>
> def func(i: => Int): Unit = {
> lazy val cached = {
> println("Calculating")
> i * i
> }
> cached
> }
>
> And then made the following calls:
>
> func(3)
> func(3)
>
> that my lazy evaluation should only happen once? - as the function is
> getting called with the same parameter the second time. But this is not the
> case when I test. Could someone explain what I am missing here?
>
> Thanks
> Des
>
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Best regards,
Brian Maso
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