BOOST_FUSION_ADAPT_STRUCT的限制 [英] Limits of BOOST_FUSION_ADAPT_STRUCT
问题描述
我尝试使用BOOST_FUSION_ADAPT_STRUCT宏，并尝试了一些幼稚的操作，例如使用Fusion来打印任意结构.
I have tried to play with the BOOST_FUSION_ADAPT_STRUCT macro and tried some naive things such as use Fusion to print any arbitrary structure.
从此示例开始文档中提供的代码，我无法在适应的结构上执行融合序列允许的某些操作.
Starting from this example code given in the documentation, I was unable to perform on my adapted structure some operations that are allowed with a fusion sequence.
#include <boost/fusion/adapted.hpp>
#include <boost/fusion/sequence/io/out.hpp>
#include <boost/fusion/sequence/intrinsic.hpp>
#include <boost/fusion/view.hpp>
#include <iostream>
namespace fuz = boost::fusion;
namespace demo
{
struct employee
{
std::string name;
int age;
};
}
// demo::employee is now a Fusion sequence
BOOST_FUSION_ADAPT_STRUCT(
demo::employee,
(std::string, name)
(int, age))
int main()
{
// tried to initialize an employee like a fusion sequence
// but it didnt work
// demo::employee e("bob", 42);
demo::employee e;
e.name = "bob";
e.age = 42;
// Access struct members with fusion random access functions
// ok
std::cout << fuz::at_c<0>(e) << std::endl;
// tried to print the struct like any othe fusion sequence
// didnt work
// std::cout << e << std::endl;
// I made it work by using a fusion view
// is it the right way?
std::cout << fuz::as_nview<0, 1>(e) << std::endl;
}
这使我想到以下问题:
-
为什么Fusion magik无法在此处运行?
Why the Fusion magik does not operate here?
使用视图是打印适应的结构的正确方法吗?
Using a view is the correct way to print an adapted struct?
经过修改的结构可以用作融合序列多远?
How far can an adapted struct be used as a Fusion sequence?
推荐答案
来自boost::fusion文档:
I/O操作符在命名空间boost :: fusion中重载
The I/O operators are overloaded in namespace boost::fusion
这意味着，如果要隐式集成这些operator<<，则需要将boost::fusion名称空间注入当前的名称空间(此处为::)，或显式使用它们.
Which means that if you want a implicit integration of these operator<<, you will need to inject the boost::fusion namespace in your current namespace (:: here), or use them explicitly.
总而言之，添加:
using namespace boost::fusion;
应该适合您的情况. 或为明确使用，您将必须编写:
Should work in your case. Or for an explicit use, you will have to write:
boost::fusion::operator<<(std::cout, e) << std::endl;
---编辑---
稍微阅读boost::fusion的代码后，您似乎感到困惑，因为 Koenig对boost::fusion::operators::operator<<的查找是在您的参数是真实的boost::fusion::sequence的情况下选择的.
After reading boost::fusion's code a bit, it seem that you are confused because of the Koenig's lookup of boost::fusion::operators::operator<< which is selected in case your argument is a real boost::fusion::sequence.
这就是为什么您无需注入boost::fusion名称空间，也无需显式调用boost::fusion::operator<<来定义boost::fusion名称空间中定义的类型的原因.
This is why you don't need to inject the boost::fusion namespace, nor explicitly call boost::fusion::operator<< for types defined in the boost::fusion namespace.
一些解释:
我不会解释Koenig查找的整个概念(也称为此处的参数依赖查找-ADL)，因为这不是重点，但基本上，它指出，如果您使用类型在名称空间内的变量，则函数查找会扩展到该参数的名称空间
I won't explain the whole concept of Koenig's lookup (also known as Argument Dependent Lookup - ADL) here since that is not the point, but basically, it states that in case you are using a variable whose type is inside a namespace, then the function lookup extends to the namespace of that parameter.
在这种特殊情况下，包括boost/fusion/sequence/io/out.hpp在内将定义boost::fusion::operator::operator<<，然后将其注入boost::fusion命名空间中.
In this particular case, including boost/fusion/sequence/io/out.hpp will define boost::fusion::operator::operator<< which will then be injected in the boost::fusion namespace.
$ cat /usr/local/include/boost/fusion/sequence/io/out.hpp
[...]
namespace boost { namespace fusion
{
[...]
namespace operators
{
template <typename Sequence>
inline typename
boost::enable_if<
fusion::traits::is_sequence<Sequence>
, std::ostream&
>::type // this is just a SFINAE trick to ensure
// the function will only be selected for
// actual boost::fusion::sequence
operator<<(std::ostream& os, Sequence const& seq)
{
return fusion::out(os, seq); // this will print out the sequence
}
}
using operators::operator<<; // here the operator<< is injected
// in boost::fusion
}}
这意味着使用 operator<< 具有类型在 boost::fusion 命名空间中的参数的调用将找到适当的重载.
使用类型不在此命名空间中的参数进行调用将无法解析operator<<的正确重载(在您的示例中就是这种情况).
Calls using arguments whose type is not located in this namespace will fail to resolve the proper overload of operator<< (this is the case in your example).
您可以通过在boost::fusion命名空间中定义类型来检查.
You can check that by defining your type in the boost::fusion namespace.
namespace boost { namespace fusion {
struct employee
{
std::string name;
int age;
};
}}
BOOST_FUSION_ADAPT_STRUCT(
boost::fusion::employee,
(std::string, name)
(int, age))
[...]
boost::fusion::employee e;
std::cout << e << std::endl; // ADL will work here
侧面说明:如果要调试这些名称查找问题，则应使用gdb.这样，您将始终知道选择了哪个过载.在这种情况下:
Side note: If you want to debug these kind of name lookup issues, you should use gdb. That way you will always know which overload was chosen. In this case:
$ cat fusion.cpp
#include <iostream>
#include <cstdlib>
#include <boost/fusion/container/vector.hpp>
#include <boost/fusion/sequence/io.hpp>
int main(int, char**)
{
boost::fusion::vector<int, char> foo(42, '?');
std::cout << foo << std::endl;
return EXIT_SUCCESS;
}
$ gdb -q ./fusion
Reading symbols for shared libraries ... done
(gdb) b 10
Breakpoint 1 at 0x1000012f7: file fusion.cpp, line 10.
(gdb) r
Starting program: /Users/avallee/Projects/tmp/fusion
Reading symbols for shared libraries ++............................. done
Breakpoint 1, main (unnamed_arg=0x7fff5fbffb60, unnamed_arg=0x7fff5fbffb60) at fusion.cpp:10
10 std::cout << foo << std::endl;
(gdb) s
boost::fusion::operators::operator<< <boost::fusion::vector<int, char, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_> > (os=@0x7fff762b5f10, seq=@0x7fff5fbffb18) at out.hpp:38
38 return fusion::out(os, seq);
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