gcc compiler warning: “will be initialized after/when initialized here”
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From: http://blog.csdn.net/qtyl1988/article/details/6818325
先看看我为二叉树写的结点类:
template <class ElemType>class TreeNode{private: TreeNode* lchild; TreeNode* rchild;public: ElemType data; TreeNode(const ElemType& item,TreeNode* lptr=NULL,TreeNode* rptr=NULL): data(item),lchild(lptr),rchild(rptr){ } ~TreeNode(){ } TreeNode*& LeftChild(){return lchild;} TreeNode*& RightChild(){return rchild;}};
好了,现在单看其中的构造函数:
TreeNode(const ElemType& item,TreeNode* lptr=NULL,TreeNode* rptr=NULL): data(item),lchild(lptr),rchild(rptr){ }
有问题吗?有!编译一下,立马出现warning,全文记不清了,但是肯定有这么两个短语:"will be initialized after/when initialized here"。啥意思?其实我也没理解。然后上网搜,答案出来了,你需要把这个构造函数的初始化列表改一改:
TreeNode(const ElemType& item,TreeNode* lptr=NULL,TreeNode* rptr=NULL): lchild(lptr),rchild(rptr),data(item){ }
其实gcc给的这个warning就是为了提醒你,初始化列表中数据成员的初始化顺序应该和你在类的声明中给出数据成员的顺序一致!用英语说,就是“Make sure the members appear in the initializer list in the same order as they appear in the class!”(这句话来自我找到答案的那个网站。)
当然,你还可以不使用初始化列表,而把初始化数据成员的工作放在构造函数体内。不过或许你需要为此重载赋值操作符“=”。
话又说回来,为什么初始化列表的顺序还有讲究?为什么会这样?
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