我想创建一个类似于std :: cout的类。我知道如何重载>>和<<运营商,但我想重载<<运算符所以它将是输入,就像在std :: cout中一样。
应该是这样的:
class MyClass
{
std::string mybuff;
public:
//friend std::???? operator<<(????????, MyClass& myclass)
{
}
}
.
.
.
MyClass class;
class << "this should be stored in my class" << "concatenated with this" << 2 << "(too)";
由于
答案 0 :(得分:4)
class MyClass
{
std::string mybuff;
public:
//replace Whatever with what you need
MyClass& operator << (const Whatever& whatever)
{
//logic
return *this;
}
//example:
MyClass& operator << (const char* whatever)
{
//logic
return *this;
}
MyClass& operator << (int whatever)
{
//logic
return *this;
}
};
答案 1 :(得分:0)
我认为最常见的答案是:
class MyClass
{
std::string mybuff;
public:
template<class Any>
MyClass& operator<<(const Any& s)
{
std::stringstream strm;
strm << s;
mybuff += strm.str();
}
MyClass& operator<<( std::ostream&(*f)(std::ostream&) )
{
if( f == std::endl )
{
mybuff +='\n';
}
return *this;
}
}
std :: endl是从Timbo的回答here
中粘贴的感谢您的回答!