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As mentioned in ticket #24346, scalar() should not change lvalueness. The fact that it did was a side effect of the implementation and a bug. foo(scalar substr(....)) should pass a substr lvalue to foo just as it would without scalar() or with a $ prototype (which is meant to be equivalent to scalar()). This also makes it possible to force scalar context in list assignment to lvalue subroutines, as in (foo(), scalar bar()) = @list.
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Father Chrysostomos
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Dec 13, 2011
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