The implicit conversion to immutable is only possible inside strongly pure
functions. When the parameter is 'in int[]' foo cannot be strongly pure,
only const pure. As 'in int[2]' is a value type, the second foo can be
strongly pure.
'new' expressions will hopefully be able to be converted to immutable
eventually, along with array concatenation and array.dup.
It is also likely that the following will be valid code (const pure foo
called with immutable arguments)
int[] foo(in int[] x) pure {
return new int[1];
}
void main() {
immutable x = foo([1, 2, 3].idup);
}
"bearophile" <bearophileh...@lycos.com> wrote in message
news:j6iom9$2g1m$1...@digitalmars.com...
> Do you know why this program doesn't compile (with DMD 2.056head)?
>
>
> immutable(int[]) foo(in int[] x) pure {
> return new int[1];
> }
> void main() {}
>
>
> It gives:
> test.d(2): Error: cannot implicitly convert expression (new int[](1u)) of
> type int[] to immutable(int[])
>
> This program instead compiles (now x is int[2] instead of int[]):
>
> immutable(int[]) foo(in int[2] x) pure {
> return new int[1];
> }
> void main() {}
>
>
> Bye and thank you,
> bearophile
