[英]Inherit from std::ostringstream
In my project I use a class class called Message which inherits from std::ostringstream to print out human readable information of other class types. 在我的项目中,我使用一个名为Message的类,它继承自std :: ostringstream,打印出其他类类型的人类可读信息。
So its << operator is overloaded several times taking my own class types which I want to print out. 所以它的<<运算符多次重载我自己的类类型,我想打印出来。
class Message : public ostringstream
{
public:
Message() {};
virtual ~Message() {};
Message& operator <<(const MyTypeA &a) { return *this << a.getStr(); };
Message& operator <<(const MyTypeB &b) { return *this << b.identifier(); };
Message& operator <<(const MyTypeC &c) { return *this << c.foo(); };
// Pass everything unknown down to ostringstream, and always return a Message&
template<class T>
Message& operator <<(const T &t)
{
(std::ostringstream&)(*this) << t;
return *this;
}
};
Without the template 没有模板
MyTypeA a,b;
Message m;
m << a << "-" << b;
the code would not compile, as (m << a << "-") would return a ostringstream& which would not be able to take 'b'. 代码不会编译,因为(m << a <<“ - ”)将返回一个ostringstream并且无法取'b'。 So I use the template to make sure to always return a Message&. 所以我使用模板确保始终返回Message&。
My Problem: 我的问题:
Message m;
m << "Hello, world" << std::endl;
generates a veeery long compiler error and I do not know why. 生成一个veeery长编译器错误,我不知道为什么。
Here is a minimalistic "not" compilable example of my problem, and here is the corresponding compiler output. 这是我的问题的简约“不”可编译示例, 这里是相应的编译器输出。
Do not derive from any of the stream classes to create a new stream! 不要从任何流类派生来创建一个新的流! The only reason to derive from std::ostream
or std::istream
is to create a stream properly set up to use a suitable stream buffer. 从std::ostream
或std::istream
派生的唯一原因是创建一个正确设置的流以使用合适的流缓冲区。 Instead of deriving from a stream, you should derive from std::streambuf
to create a new source or destination. 您应该从std::streambuf
派生来创建新的源或目标,而不是从std::streambuf
。 To create output operators for new types you'd overload operator<<()
with std::ostream&
as first argument and your custom type as second argument. 要为新类型创建输出运算符,您需要使用std::ostream&
作为第一个参数并将自定义类型作为第二个参数重载operator<<()
。 Likewise with std::istream&
. 与std::istream&
。
BTW, the problem you encountered is because std::endl
is a function template. 顺便说一句,你遇到的问题是因为std::endl
是一个函数模板。 Trying to pass a function template somewhere requires that the appropriate instantiation can be deduced. 尝试在某处传递函数模板需要可以推导出适当的实例化。 Since your output operators don't cover a suitable signature, the instantiation of std::endl
can't be deduced. 由于输出运算符不包含合适的签名,因此无法推导出std::endl
的实例化。 I could state what's needed but that would be pointless as it is the wrong way to go anyway. 我可以说明需要什么,但这是毫无意义的,因为无论如何都是错误的方式。
Here is how to make your code compile: http://pastebin.com/xAt5junf 以下是如何编译代码: http : //pastebin.com/xAt5junf
The relevant excerpt: 相关摘录:
class Message : public ostringstream
{
public:
Message() {};
virtual ~Message() {};
};
ostream& operator <<(ostream &os, const MyTypeA &a) {
return os << a.getStr();
};
ostream& operator <<(ostream &os, const MyTypeB &b) {
return os << b.getStr();
};
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