On Tue, Mar 27, 2012 at 3:46 PM, Igor R
I thought its purpose was allow binding, but not evaluating, of a function object.
...when passed to bind() as a not first argument.
Ah, right, thx.
So that this should work.
#include <iostream> #include
#include void f( int i ) { std::cout << i << std::endl; }
template <typename NullaryCallable> void g( const NullaryCallable & callable ) { callable( ); }
int main( ) { g( protect( bind( f, _1 ) )( 1 ) ); }
But it doesn't, and instead thinks I'm calling g() with the return type of f(), just as if I'd not used protect at all.
Sure, protect doesn't do anything here.
Is it possible to have bind generate a fully bound, nullary callable object?
bind(f, 1);
Yeah, but that doesn't quite fit the bill! Ultimately I'm doing this for each member of a range, via std algorithm, so I need to bind a unary argument to produce a nullary function object, which I can then pass to an evaluation context, ie my g( ) function. If I've generated the function object with bind and placeholder(s), it seems there's no way to prevent the binding of the last placeholder from triggering evaluation, when what I require is an unevaluated nullary function object. Thx, Rob.