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

From: Stuart Dootson (stuart.dootson_at_[hidden])
Date: 2006-01-24 11:58:05

On 1/24/06, David A. Greene <greened_at_[hidden]> 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