1 Aug
2007
1 Aug
'07
3:51 a.m.
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...