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From: Stefan Seefeld (seefeld_at_[hidden])
Date: 2007-07-31 23:51:24
Eric Niebler wrote:
> I think the issue is that, although the literal "0" can be implicitly
> converted to a pointer, when passed to std::fill(), its type is deduced
> to be "int", at which point it can no longer be converted to a pointer.
I see ! I hadn't realized that the third parameter type was deduced from
the argument. (Instead I was fixated on the use of 'NULL' instead of '0'.)
Thanks,
Stefan
-- ...ich hab' noch einen Koffer in Berlin...
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