C++ 我如何使用 cout << myclass
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How can I use cout << myclass
提问by ali
myclassis a C++ class written by me and when I write:
myclass是我写的一个 C++ 类，当我写的时候：
myclass x;
cout << x;
How do I output 10or 20.2, like an integeror a floatvalue?
我如何输出10or 20.2，比如一个integer或一个float值？
回答by Jerry Coffin
Typically by overloading operator<<for your class:
通常通过operator<<为您的类重载：
struct myclass {
int i;
};
std::ostream &operator<<(std::ostream &os, myclass const &m) {
return os << m.i;
}
int main() {
myclass x(10);
std::cout << x;
return 0;
}
回答by Rich Adams
You need to overload the <<operator,
您需要重载<<运算符，
std::ostream& operator<<(std::ostream& os, const myclass& obj)
{
os << obj.somevalue;
return os;
}
Then when you do cout << x(where xis of type myclassin your case), it would output whatever you've told it to in the method. In the case of the example above it would be the x.somevaluemember.
然后，当您执行此操作时cout << x（您的情况中x的类型myclass为where ），它将输出您在方法中告诉它的任何内容。在上面的例子中，它将是x.somevalue成员。
If the type of the member can't be added directly to an ostream, then you would need to overload the <<operator for that type also, using the same method as above.
如果成员的类型不能直接添加到 an ostream，那么您还需要<<使用与上述相同的方法重载该类型的运算符。
回答by Tristram Gr?bener
it's very easy, just implement?:
这很容易，只需实现？：
std::ostream & operator<<(std::ostream & os, const myclass & foo)
{
os << foo.var;
return os;
}
You need to return a reference to os in order to chain the outpout (cout << foo << 42 << endl)
您需要返回对 os 的引用以链接输出（cout << foo << 42 << endl）
回答by Emil Mocan
Alternative:
选择：
struct myclass {
int i;
inline operator int() const
{
return i;
}
};
