Question

我正在尝试使用boost :: bind将curried函数链接在一起，并获得无法解析的编译器错误。我可以做的最简单的例子，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);
}

因此错误而失败：

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

Answer 1

在这种情况下，您需要使用boost::protect()，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的表达式，而是希望嵌套的绑定表达式作为参数传递给外部绑定函数。这将是：

boost::bind(&A::g<F>, this, boost::protect(fn), ::_1)(c);

boost :: bind成员函数 - 部分应用程序链接

1 个答案: