Hi John Mercier !
Thank you for you comment :)
I didn't know about Function.identity(), and, since I like to have the more explicit code possible, I tried it, but I obtained an error.
Function.identity()
error: incompatible types: no instance(s) of type variable(s) T exist so that Function<T,T> conforms to ToIntFunction<? super Integer>
After some googling, I founded this stackoverflow thread and replace the i->i by Integer::intValue instead, which worked too.
i->i
Integer::intValue
Nice! I think identity() returns an Integer in this case and mapToInt expects an int. There is no unboxing going on here. Sorry for the confusion.
identity()
mapToInt
int
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Hi John Mercier !
Thank you for you comment :)
I didn't know about
Function.identity(), and, since I like to have the more explicit code possible, I tried it, but I obtained an error.After some googling, I founded this stackoverflow thread and replace the
i->ibyInteger::intValueinstead, which worked too.Nice! I think
identity()returns an Integer in this case andmapToIntexpects anint. There is no unboxing going on here. Sorry for the confusion.