boost :: bind/boost :: function自從1.55

Question

奇怪的行爲變化我試圖編譯a project是針對Boost 1.55針對較新的Boost 1.63編寫的，而且我遇到了與bind/function相關的非常奇怪的錯誤。下面是完整的，簡單的測試用例：

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

template < typename Arg1 = int, typename Arg2 = int, typename Arg3 = int > 
class foo 
{ 
public: 
    using function_t = boost::function3< void, Arg1, Arg2, Arg3 >; 

    void set_function(function_t f) 
    { 
    func_ = f; 
    } 

private: 
    function_t func_; 
}; 

class bar 
{ 
public: 
    bar() 
    { 
    foo_.set_function(boost::bind(&bar::func, this, _1, _2)); 
    } 

private: 
    void func(int const&, int&) {} 

    foo< int, int > foo_; 
}; 

int main() 
{ 
    bar x; 
    return 0; 
} 

...和錯誤的一些選擇片段我越來越：

/usr/include/boost/bind/bind.hpp:398:35: error: no match for call to ‘(boost::_mfi::mf2<void, bar, const int&, int&>) (bar*&, int, int)’ 
/usr/include/boost/bind/mem_fn_template.hpp:278:7: note: candidate: R boost::_mfi::mf2<R, T, A1, A2>::operator()(T*, A1, A2) const [with R = void; T = bar; A1 = const int&; A2 = int&] <near match> 
/usr/include/boost/bind/mem_fn_template.hpp:278:7: note: conversion of argument 3 would be ill-formed: 
/usr/include/boost/bind/bind.hpp:398:35: error: cannot bind non-const lvalue reference of type ‘int&’ to an rvalue of type ‘int’ 
/usr/include/boost/bind/mem_fn_template.hpp:283:25: note: candidate: template<class U> R boost::_mfi::mf2<R, T, A1, A2>::operator()(U&, A1, A2) const [with U = U; R = void; T = bar; A1 = const int&; A2 = int&] 
/usr/include/boost/bind/mem_fn_template.hpp:283:25: note: template argument deduction/substitution failed: 
/usr/include/boost/bind/bind.hpp:398:35: note: cannot convert ‘(& a)->boost::_bi::rrlist3<int, int, int>::operator[](boost::_bi::storage3<boost::_bi::value<bar*>, boost::arg<1>, boost::arg<2> >::a3_)’ (type ‘int’) to type ‘int&’ 
/usr/include/boost/bind/mem_fn_template.hpp:291:25: note: candidate: template<class U> R boost::_mfi::mf2<R, T, A1, A2>::operator()(const U&, A1, A2) const [with U = U; R = void; T = bar; A1 = const int&; A2 = int&] 
/usr/include/boost/bind/mem_fn_template.hpp:291:25: note: template argument deduction/substitution failed: 
/usr/include/boost/bind/bind.hpp:398:35: note: cannot convert ‘(& a)->boost::_bi::rrlist3<int, int, int>::operator[](boost::_bi::storage3<boost::_bi::value<bar*>, boost::arg<1>, boost::arg<2> >::a3_)’ (type ‘int’) to type ‘int&’ 
/usr/include/boost/bind/mem_fn_template.hpp:299:7: note: candidate: R boost::_mfi::mf2<R, T, A1, A2>::operator()(T&, A1, A2) const [with R = void; T = bar; A1 = const int&; A2 = int&] 
/usr/include/boost/bind/mem_fn_template.hpp:299:7: note: no known conversion for argument 1 from ‘bar*’ to ‘bar&’ 

如果我嘗試具有升壓1.55另一臺機器上，它編譯正好。 （當指向構建Boost 1.55時，該項目在同一臺機器上編譯也沒問題，所以問題看起來並不是編譯器。）所以，顯然Boost中的某些內容已經改變，導致它失敗。

我沒有寫這段代碼，我不是很熟悉boost::function和boost::bind的膽量。如果有人可以向我解釋爲什麼這會破壞新的Boost，或者a）如何修復它，或者b）爲什麼上面的代碼被破壞，這將是非常感謝！

Answer 1

問題是boost::function在匹配函數簽名（我認爲更接近地匹配std::function的行爲）方面變得更加嚴格。您的函數聲明爲boost::function3< void, Arg1, Arg2, Arg3 >，但是，您綁定的函數需要int const&和int&。

有幾種方法可以解決這個問題。你可以：

1. 變化foo<int, int>在bar的instantion爲foo<int const&, int&>。
2. 將function_t的聲明更改爲boost::function3<void, Arg1, Arg2 const&, Arg3&>。
3. 將bar::func的簽名更改爲通過值接受其參數。