Templates and g++ 4.7

Templates and g++ 4.7

本文关键字：g++ and Templates      更新时间：2023-10-16

我有这个代码

namespace MSL{
template <typename T> class TListNode;
template <typename T> class TList;
...

和

template <typename T> 
int TList<T>::add(T v) {
TListNode<T>  *pn;  
pn = new TListNode<T>;
...

和 

class TMergeNode {
 public:
 inline      TMergeNode(int cluster1=-1, int cluster2=-1, TCMData mergeVal=0);
 inline      TMergeNode(TMergeNode &b);                // copy constructor
 ...

它在旧版本的 g++ 中编译正常，但现在使用 4.7 版，我收到以下错误：

./msl/MSL_List_Template.h: In instantiation of ‘int MSL::TList<T>::add(T) [with T = TMergeNode]’:
clustermerges.cpp:282:33:   required from here
./msl/MSL_List_Template.h:616:23: error: no matching function for call to ‘TMergeNode::TMergeNode(TMergeNode)’
./msl/MSL_List_Template.h:616:23: note: candidates are:
In file included from main.cpp:78:0:
clustermerges.cpp:70:8: note: TMergeNode::TMergeNode(TMergeNode&)
clustermerges.cpp:70:8: note:   no known conversion for argument 1 from ‘TMergeNode’ to ‘TMergeNode&’
clustermerges.cpp:68:8: note: TMergeNode::TMergeNode(int, int, MSL::TCMData)
clustermerges.cpp:68:8: note:   no known conversion for argument 1 from ‘TMergeNode’ to ‘int’

任何想法将不胜感激

在您的代码中，您正在尝试将临时引用绑定到非常量引用。这是不允许的。

复制构造函数的正确签名是：

class TMergeNode {
public:
 inline      TMergeNode(const TMergeNode &b);                // copy constructor
 //                     ^^^^^
相关文章：