On Monday, 20 July 2015 at 15:12:28 UTC, Marc Schütz wrote:
On Monday, 20 July 2015 at 15:08:16 UTC, Nicholas Wilson wrote:
But the lambda takes a ref parameter...

Yes, but it never writes to it:

    x.each!((ref a) => a + 1);

Instead, this should work:

    x.each!((ref a) => a = a + 1);

... as a short-hand for:

    x.each!((ref a) { a = a + 1; });

This works! Thanks.

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