Re: [Boost-users] [lambda] "Invalid use of void expression"

24 Jan 2006

      On 1/24/06, David A. Greene <greened@obbligato.org> wrote:
...
I'm having some trouble with BLL and Boost.Function in this simple
testcase:
#include <boost/lambda/bind.hpp>
#include <boost/function.hpp>
#include <iostream>
class Test {
public:
   void foo(int a) { std::cout << "a is " << a << std::endl; };
};
template<typename Function>
class Caller {
private:
   Function func;
public:
   Caller(Function f) : func(f) {};
void bar(void) { func(); };
};
int main(void)
{
   Test tester;
   int value = 10;
Caller<boost::function<void (void)> >
printer(boost::lambda::bind(boost::lambda::bind(&Test::foo, tester,
boost::lambda::_1), value));
printer.bar();
return(0);
}
The g++ 4.0.3 compiler complains (full trace below):
/usr/include/boost/lambda/detail/lambda_functor_base.hpp:408: error:
invalid use of void expression
Is it not legal to bind a lambda-bound function?
-Dave
It's probably more that it makes the compilers job too
difficult...anyway - I think you'll find that one of these two does
what you want (I've assumed a 'using namespace boost::lambda;' because
it declutters the code):

   Caller<boost::function<void ()> > printer1(bind(&Test::foo, &tester, value));
   Caller<boost::function<void ()> > printer1(bind(&Test::foo,
&tester, boost::ref(value)));

With printer1, the value of 'value' is baked into the bound object.
With printer2, a reference to 'value' is baked into the bound object.
This means that if you change the value of 'value', printer1 won't
reflect the change, but printer2 will - the following code

   printer1.bar();
   printer2.bar();
   value = 1234;
   printer1.bar();
   printer2.bar();

will give the output

   10
   10
   10
   1234

HTH!
Stuart Dootson