我的问题很简单,但是我陷入了困境。如何从基类中选择所需的构造函数?
// node.h
#ifndef NODE_H
#define NODE_H
#include <vector>
// definition of an exception-class
class WrongBoundsException
{
};
class Node
{
public:
...
Node(double, double, std::vector<double>&) throw (WrongBoundsException);
...
};
#endif
// InternalNode.h
#ifndef INTERNALNODE_H
#define INTERNALNODE_H
#include <vector>
#include "Node.h"
class InternalNode : public Node
{
public:
// the position of the leftmost child (child left)
int left_child;
// the position of the parent
int parent;
InternalNode(double, double, std::vector<double>&, int parent, int left_child) throw (WrongBoundsException);
private:
int abcd;
};
#endif
// InternalNode.cpp
#include "InternalNode.h"
#define UNDEFINED_CHILD -1
#define ROOT -1
// Here is the problem
InternalNode::InternalNode(double a, double b, std::vector<double> &v, int par, int lc)
throw (WrongBoundsException)
: Node(a, b, v), parent(par), left_child(lc)
{
std::cout << par << std::endl;
}
我得到:
$ g++ InternalNode.cpp
InternalNode.cpp:16:错误:“InternalNode::InternalNode(double,double,std::vector>&,int,int)的声明(WrongBoundsException)”引发了不同的异常
InternalNode.h:17:错误:来自先前的声明“InternalNode::InternalNode(double,double,std::vector>&,int,int)”
更新0:修复丢失:
更新1:修复抛出异常
最佳答案
此简化的代码可以正确编译,但是由于缺少基类的构造函数定义,因此不会链接:
#include <vector>
// definition of an exception-class
class WrongBoundsException {
};
class Node {
public:
Node(double, double, std::vector<double>&)
throw (WrongBoundsException);
};
class InternalNode : public Node {
public:
// the position of the leftmost child (child left)
int left_child;
// the position of the parent
int parent;
InternalNode(double, double, std::vector<double>&,
int parent, int left_child)
throw (WrongBoundsException);
private:
int abcd;
};
// Note added exception specification
InternalNode::InternalNode(double a, double b,
std::vector<double> &v,
int par, int lc) throw (WrongBoundsException)
: Node(a, b, v), parent(par), left_child(lc)
{
}
顺便说一句,为什么您觉得需要使用异常规范?在C++中,它们通常看起来有点浪费时间。