[英]Overloading operator<<: cannot bind lvalue to ‘std::basic_ostream<char>&&’
I have a class that uses a nested class, and want to use the nested class operator<<
to define operator<<
in the upper class. 我有一个使用嵌套类的类,并希望使用嵌套类
operator<<
operator<<
在上层类中定义operator<<
。 Here is how my code looks like: 以下是我的代码的样子:
#include <memory>
#include <iostream>
template<typename T>
struct classA {
struct classB
{
template<typename U>
friend inline std::ostream& operator<< (std::ostream &out,
const typename classA<U>::classB &b);
};
classB root;
template<typename U>
friend std::ostream& operator<< (std::ostream &out,
const classA<U> &tree);
};
template<typename T>
inline std::ostream& operator<< (std::ostream &out,
const classA<T> &tree)
{
out << tree.root;
return out;
}
template<typename T>
inline std::ostream& operator<< (std::ostream &out,
const typename classA<T>::classB &b)
{
return out;
}
int main()
{
classA<int> a;
std::cout << a;
}
When compiling without support for C++11, the definition of operator<< for the inner class seems not to be found by the compiler: 在不支持C ++ 11的情况下进行编译时,编译器似乎找不到内部类的operator <<的定义:
so.hpp:24:7: error: no match for 'operator<<' in 'out << tree.classA<int>::root' so.hpp:24:7: note: candidates are: ...
With GCC 4.6 and 4.7 when compiling with std=c++0x: 使用std = c ++ 0x进行编译时使用GCC 4.6和4.7:
so.hpp:21:3: error: cannot bind 'std::ostream {aka std::basic_ostream<char>}' lvalue to 'std::basic_ostream<char>&&' In file included from /usr/include/c++/4.7/iostream:40:0, from so.hpp:2: /usr/include/c++/4.7/ostream:600:5: error: initializing argument 1 of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = classA<int>::classB]'
Can someone tell me why this code is not legal, and what's the best way to do what I want? 有人能告诉我为什么这段代码不合法,什么是做我想要的最好的方法?
You have a problem with a "non-deducible context" in this operator 您在此运算符中遇到“不可导入的上下文”问题
template<typename T>
inline std::ostream& operator<< (std::ostream &out,
const typename classA<T>::classB &b)
{
return out;
}
The compiler cannot figure out what values of T
will result in a classB
that matches the parameter you want to pass. 编译器无法确定哪个
T
值会导致classB
与您要传递的参数匹配。 So this template is not considered! 所以这个模板不予考虑!
In C++11 mode, the compiler then goes on to find a close match from the standard library 在C ++ 11模式下,编译器继续从标准库中找到一个紧密匹配
operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&)
where it can match _Tp
to just about any type, including classA<T>::classB
, but notes that the first parameter doesn't match. 它可以匹配
_Tp
几乎任何类型,包括classA<T>::classB
,但注意第一个参数不匹配。
Bo provided the reason why this is happening (the type T
is not deducible in the call to the nested operator<<
. A simple workaround for this, and something that I recommend in general, not only here, is not befriending a template, but rather a single free function. For that you will need to define the function inline: Bo提供了为什么会发生这种情况的原因 (类型
T
在嵌套operator<<
的调用中是不可推断的。这是一个简单的解决方法,我推荐的一般情况,不仅仅是在这里,不是模板的朋友,而是而是一个单独的自由函数。为此你需要定义内联函数:
template<typename T>
struct classA {
struct classB
{
friend inline std::ostream& operator<< (std::ostream &out,
const classB &b) {
// definition goes here
}
};
classB root;
friend std::ostream& operator<< (std::ostream &out,
const classA<U> &tree) {
// definition goes here
}
};
There are a couple of differences among the two approaches. 这两种方法有两点不同。 The most important one is that this approach will have the compiler define a non-templated overload for
operator<<
for each instantiation of the template, which because it is no longer a template, does not depend on deducing the arguments. 最重要的一点是,这种方法将使编译器为模板的每个实例化定义
operator<<
的非模板化重载,因为它不再是模板,不依赖于推导参数。 Another side effects are that the approach is a little tighter (you are only befriending one function, while in your initial approach you befriended the template and all possible instantiations (which can be used as a loophole to gain access to your class internals). Finally the functions so defined will only be found through ADL, so there are less overloads of operator<<
for the compiler to consider when the argument is not ClassA<T>
or ClassA<T>::ClassB
. 另一个副作用是方法更紧凑 (你只是与一个函数交朋友,而在你最初的方法中,你结识了模板和所有可能的实例化(可以作为漏洞来获得对你的类内部的访问)。这样定义的函数只能通过ADL找到,因此当参数不是
ClassA<T>
或ClassA<T>::ClassB
时,编译器要考虑的operator<<
重载次数较少。
How access can be gained with your approach 如何通过您的方法获得访问权限
namespace {
struct intruder {
ClassA & ref;
intruder( ClassA& r ) : ref(r) {}
};
template <>
std::ostream& operator<< <intruder>( std::ostream& _, ClassA<intruder> const& i ) {
std::cout << i.ref.private_member << std::endl;
return _;
}
}
Alternative 替代
Alternatively you can befriend a particular specialization of a template. 或者,您可以与模板的特定专业化建立联系。 That will solve the
intruder
problem, as it will only be open to operator<<
to ClassA<intruder>
, which has a much lesser impact. 这将解决
intruder
问题,因为它只对operator<<
to ClassA<intruder>
,后者的影响要小得多。 But this will not solve your particular issue, as the type would still not be deducible. 但这不会解决您的特定问题,因为该类型仍然无法推断。
Try this: 试试这个:
template<typename T>
inline std::ostream& operator<< (std::ostream &out,
const classA<T> &tree)
{
//out << tree.root;
::operator<<( out, tree.root);
return out;
}
and then you will get a straightforward confession of ineptitude: 然后你会得到一个直率的无能告白:
test.cpp:34:3: error: no matching function for call to ‘operator<<(std::ostream&, const classA<int>::classB&)’
test.cpp:34:3: note: candidates are:
test.cpp:23:22: note: template<class T> std::ostream& operator<<(std::ostream&, const typename classA<T>::classB&)
test.cpp:30:22: note: template<class T> std::ostream& operator<<(std::ostream&, const classA<T>&)
Workaround: maybe you can use a member function in nested classB, and use it instead of operator<< ... Of course, that solution has a multitude of drawbacks, but it may get you out of this hurry. 解决方法:也许您可以在嵌套的classB中使用成员函数,并使用它而不是运算符<< ...当然,该解决方案有许多缺点,但它可能会让您摆脱这种匆忙。
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