#include <iostream>
#include <sstream>
class Vector
{
double _x;
double _y;
public:
Vector(double x,double y) : _x(x),_y(y) {}
double getX() { return _x; }
double getY() { return _y; }
operator const char*()
{
std::ostringstream os;
os << "Vector(" << getX() << "," << getY() << ")";
return os.str().c_str();
}
};
int main()
{
Vector w1(1.1,2.2);
Vector w2(3.3,4.4);
std::cout << "Vector w1(" << w1.getX() << ","<< w1.getY() << ")"<< std::endl;
std::cout << "Vector w2(" << w2.getX() << ","<< w2.getY() << ")"<< std::endl;
const char* n1 = w1;
const char* n2 = w2;
std::cout << n1 << std::endl;
std::cout << n2 << std::endl;
}
该计划的输出：
$./a.out Vector w1(1.1,2.2) Vector w2(3.3,4.4) Vector(3.3,4.4)
我不明白为什么我得到输出.似乎“const char * n2 = w2;”覆盖n1然后我得到两次“Vector(3.3,4.4)”.有人可以解释一下这种现象吗？
解决方法
这是未定义的行为,有时有效(运气),有时不行.
您正在返回指向临时本地对象的指针.指向临时本地对象的指针是通过调用os.str().c_str()获得的字符串对象的内部.
如果你想通过cout轻松打印这些对象,你可以重载operator<<用于输出流.喜欢：
ostream& operator<<(ostream& out,const Vector &a)
{
std::ostringstream os;
os << "Vector(" << a.getX() << "," << a.getY() << ")";
out << os.str();
return out;
}
然后
std::cout << w1 << std::endl; std::cout << w2 << std::endl;
