3

我正在尝试使用 boost::bind 将 curried 函数链接在一起，并得到我无法解决的编译器错误。我可以制作的最简单的示例无法编译：

#include <iostream>
#include <boost/bind.hpp>

class A
{
public:
    template <typename F>
    void g(F fn, char c)
    {
        fn(c);
    }

    void h(char c)
    {
        std::cout << c << std::endl;
    }

    template <typename F>
    void f(F fn, char c)
    {
        boost::bind(&A::g<F>, this, fn, ::_1)(c);
    }
};

int main(int argc, char** argv)
{
    char c = 'a';
    A a;
    a.f(boost::bind(&A::h, &a, ::_1), c);
}

失败并出现此错误：

In file included from /usr/include/boost/bind.hpp:22,
                 from test8.cpp:2:
/usr/include/boost/bind/bind.hpp: In member function ‘void boost::_bi::list3<A1, A2, A3>::operator()(boost::_bi::type<void>, F&, A&, int) [with F = boost::_mfi::mf2<void, A, boost::_bi::bind_t<void, boost::_mfi::mf1<void, A, char>, boost::_bi::list2<boost::_bi::value<A*>, boost::arg<1> > >, char>, A = boost::_bi::list1<char&>, A1 = boost::_bi::value<A*>, A2 = boost::_bi::bind_t<void, boost::_mfi::mf1<void, A, char>, boost::_bi::list2<boost::_bi::value<A*>, boost::arg<1> > >, A3 = boost::arg<1>]’:
/usr/include/boost/bind/bind_template.hpp:32:   instantiated from ‘typename boost::_bi::result_traits<R, F>::type boost::_bi::bind_t<R, F, L>::operator()(A1&) [with A1 = char, R = void, F = boost::_mfi::mf2<void, A, boost::_bi::bind_t<void, boost::_mfi::mf1<void, A, char>, boost::_bi::list2<boost::_bi::value<A*>, boost::arg<1> > >, char>, L = boost::_bi::list3<boost::_bi::value<A*>, boost::_bi::bind_t<void, boost::_mfi::mf1<void, A, char>, boost::_bi::list2<boost::_bi::value<A*>, boost::arg<1> > >, boost::arg<1> >]’
test8.cpp:21:   instantiated from ‘void A::f(F, char) [with F = boost::_bi::bind_t<void, boost::_mfi::mf1<void, A, char>, boost::_bi::list2<boost::_bi::value<A*>, boost::arg<1> > >]’
test8.cpp:29:   instantiated from here
/usr/include/boost/bind/bind.hpp:385: error: invalid use of void expression

这是使用

$ g++ --version
g++ (Ubuntu 4.4.3-4ubuntu5.1) 4.4.3
4

1 回答 1

6

您需要boost::protect()在这种情况下使用，当您将绑定表达式传递给另一个绑定调用但不希望在评估外部绑定函数的参数期间调用嵌套表达式，而是希望传递嵌套的绑定表达式原样作为外部边界函数的参数。这将是：

boost::bind(&A::g<F>, this, boost::protect(fn), ::_1)(c);
于 2012-10-02T11:22:36.343 回答