__traits(compiles...) does not call your function so it is not evaluate
twice only once, so there is no need to use memoize
On Fri, Nov 23, 2018 at 11:35 AM John Chapman via Digitalmars-d-learn <
digitalmars-d-learn@puremagic.com> wrote:
> I'm doing a fair amount of repeatedly checking if a
On Friday, 23 November 2018 at 11:29:24 UTC, Nicholas Wilson
wrote:
No, std.functional.memoize uses a hashtable to cache the
runtime results of calls to expensive functions.
assuming that the example is not oversimplified and
generateFunc1 and generateFunc2 are functions, the compiler
On Friday, 23 November 2018 at 10:34:11 UTC, John Chapman wrote:
I'm doing a fair amount of repeatedly checking if a function
compiles with __traits(compiles...), executing the function if
so, erroring out if not, like this:
static if (__traits(compiles, generateFunc1())) {
return
I'm doing a fair amount of repeatedly checking if a function
compiles with __traits(compiles...), executing the function if
so, erroring out if not, like this:
static if (__traits(compiles, generateFunc1())) {
return generateFunc1();
} static if (__traits(compiles, generateFunc2())) {
