构建我自己的类，其行为类似于输出字符串流 [英] build my own class that behaves like an output string stream
问题描述
Hello Everybody，
我希望我自己的班级表现得像下面的oss对象;
------- -----------------代码开始-----
int main（）{
int i = 67;
ostringstream oss;
oss<< " A" << i<< endl;
}
------------------------代码结束----- -
这意味着我希望能够在我的班级中使用''<<'''运算符，当然在课程中我希望能够访问传递的流。小心，我不是在找一个类似于输出的课程
cout<< one_instance_of_my_class。
要做到这一点，我想以下列方式重载''<<<'''运算符：
--- ---------------------代码开始-----
类表：public ostringstream
{
public：
friend Table&
operator<<（Table& tab，ostringstream& os）{
cout<< 这是装载的： << os.str（）<< endl;
tab.content + = os.str（）;
返回标签;
}
私人：
字符串内容;
}
------------------------代码结束------
用途如下：
------------------------代码开始-----
int main（）{
int i = 67;
表mytab;
mytab<< " A" << i<< endl;
}
------------------------代码结束----- -
如果我保留基类ostringstream，则不会执行重载的运算符，因此我无法访问传递的流。
如果省略基类，例如通过的A表示在''<<'''运算符的使用中不被视为字符串流，而是字符串。
我有意识到这可能是一个微不足道的c ++问题，但首先，我正在寻找一段时间，并没有找到任何方法来做到这一点，其次，我是新的c ++： - /。
提前谢谢非常感谢您的帮助。
VBR
gato
Hello Everybody,
I want my own class to behave like the following oss object;
------------------------ code begin -----
int main () {
int i = 67;
ostringstream oss;
oss << "A" << i << endl;
}
------------------------ code end ------
which means that i want to be able to use the ''<<'' operator with my class, and of course in the class i would like to have access to the passed stream. Caution, i''m not looking for a class which is outputstreamable like
cout << one_instance_of_my_class.
To do this i thought overloading the ''<<'' operator in the following way:
------------------------ code begin -----
class Table : public ostringstream
{
public:
friend Table&
operator<<(Table& tab, ostringstream& os) {
cout << "this is beeing loaded: " << os.str() << endl;
tab.content += os.str();
return tab ;
}
private:
string content;
}
------------------------ code end ------
to be uses like :
------------------------ code begin -----
int main () {
int i = 67;
Table mytab;
mytab << "A" << i << endl;
}
------------------------ code end ------
If i keep the base class ''ostringstream'' the overloaded operator is not executed, thus i have no access to the passed stream.
If I omit the base class, e.g. the passed "A" in the use of the ''<<'' operator is not seen as a stringstream, but as string.
I''m conscious that this might be a somehow trivial c++ question, but firstly, i was looking for a while and did not found any way to do this and secondly, i''m pretty new to c++ :-/.
Thanks in advance very much for your kind help.
VBR
gato
推荐答案
Hello Everybody，
在以下代码中，
int main（）{
ostringstream oss;
oss<< A;
}
Hello Everybody,
In the following code,
int main () {
ostringstream oss;
oss << "A";
}
所以???有什么问题？!!
So??? What''s the problem?!!
Hello Everybody，
以下代码，
int main（）{
ostringstream oss;
oss<< ; A;
}
Hello Everybody,
In the following code,
int main () {
ostringstream oss;
oss << "A";
}
你想要''为了构建你自己的接口为ostrigstream的类，对吗？
好吧，然后从运营商<<开始，他是你成功的关键。
Savage
PS：请阅读我们的发布指南..
You wan''t to build your own class that has interface as ostrigstream,right?
Well,then start from operator <<,he''s the key to your success.
Savage
PS:Please read our posting guidelines..
所以？？？有什么问题？！！
So??? What''s the problem?!!
对不起，我做了键盘错误并且发送的信息太早了。
对不起！
gato
Sorry, i did a keyboard error and sent the message too early.
Sorry for that!
gato
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