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[英]Code compiles with g++ but not clang++ if incomplete type is used in templated function
[英]C++: Templated code compiles and runs fine with clang++, but fails with g++
看一下鏈表的此實現:
#include <memory>
#include <type_traits>
#include <iostream>
using namespace std;
template<typename D>
class List {
struct Node {
shared_ptr<D> data;
Node* next;
Node(shared_ptr<D> d, Node* p, Node* n) : data(d), next(n) {}
~Node() {
data.reset();
delete next;
}
};
template <bool isconst = false>
struct iterator : public std::iterator<std::forward_iterator_tag, shared_ptr<D>> {
typedef std::forward_iterator_tag iterator_category;
typedef shared_ptr<D> value_type;
typedef std::ptrdiff_t Distance;
typedef typename conditional<isconst, const value_type&, value_type&>::type
Reference;
typedef typename conditional<isconst, const value_type*, value_type*>::type
Pointer;
typedef typename conditional<isconst, const Node*, Node*>::type
nodeptr;
iterator(nodeptr x = nullptr) : curr_node(x) {}
iterator(const iterator<false>& i) : curr_node(i.curr_node) {}
Reference operator*() const { return curr_node->data; }
Pointer operator->() const { return &(curr_node->data); }
template<bool A>
friend bool operator==(const iterator<A>& a, const iterator<A>& b) {
return a.curr_node == b.curr_node;
}
template<bool A>
friend bool operator!=(const iterator<A>& a, const iterator<A>& b) {
return !(a.curr_node == b.curr_node);
}
friend class List<D>;
iterator& operator++() {
curr_node = curr_node->next;
return *this;
}
private:
nodeptr curr_node;
};
public:
List() {
head = nullptr;
}
int len() const {
int ret = 0;
for (const auto& n : *this) {
ret++;
}
return ret;
}
~List() {
delete head;
}
std::ostream& dump(std::ostream &strm) const {
for (const auto s : *this) {
strm << *s << std::endl;
}
return strm;
}
iterator<false> begin() {
return iterator<false>(head);
}
iterator<false> end() {
return iterator<false>(nullptr);
}
iterator<true> begin() const {
return iterator<true>(head);
}
iterator<true> end() const {
return iterator<true>(nullptr);
}
private:
Node* head;
};
給我帶來問題的部分是該列表的iterator
實現。 迭代器模板應該提供可變和const
迭代器。
這是一個使用此實現的程序:
#include "List.h"
#include <iostream>
int main( int argc, const char *argv[] ) {
List<int> l;
std::cout << l.len() << std::endl;
return 0;
}
如果我使用clang++
,則程序可以編譯並運行良好,但是對於g++
,編譯失敗,並出現以下錯誤:
In file included from t.cpp:1:
List.h: In instantiation of ‘struct List<int>::iterator<false>’:
List.h:136:5: required from ‘int List<D>::len() const [with D = int]’
t.cpp:7:24: required from here
List.h:64:21: error: redefinition of ‘template<bool A> bool operator==(const List<int>::iterator<isconst>&, const List<int>::iterator<isconst>&)’
friend bool operator==(const iterator<A>& a, const iterator<A>& b) {
^~~~~~~~
List.h:64:21: note: ‘template<bool A> bool operator==(const List<int>::iterator<isconst>&, const List<int>::iterator<isconst>&)’ previously declared here
List.h:69:21: error: redefinition of ‘template<bool A> bool operator!=(const List<int>::iterator<isconst>&, const List<int>::iterator<isconst>&)’
friend bool operator!=(const iterator<A>& a, const iterator<A>& b) {
^~~~~~~~
List.h:69:21: note: ‘template<bool A> bool operator!=(const List<int>::iterator<isconst>&, const List<int>::iterator<isconst>&)’ previously declared here
此錯誤的原因是什么? 我怎樣才能解決這個問題?
問題似乎在這里:
template <bool isconst = false>
struct iterator : public std::iterator<std::forward_iterator_tag, shared_ptr<D>> {
template<bool A>
friend bool operator==(const iterator<A>& a, const iterator<A>& b) {
return a.curr_node == b.curr_node;
}
這就是說:對於isconst
所有值(外部模板參數),請定義模板函數template<bool A> bool operator==
。
因此,實例化iterator<true>
將定義template<bool A> bool operator==
,然后實例化iterator<false>
將再次定義template<bool A> bool operator==
,從而導致重新定義錯誤。
解決方案:卸下內部模板。 讓iterator
每個實例僅定義其自己的operator==
:
template <bool isconst = false>
struct iterator : public std::iterator<std::forward_iterator_tag, shared_ptr<D>> {
friend bool operator==(const iterator& a, const iterator& b) {
return a.curr_node == b.curr_node;
}
(這里的iterator
自動引用iterator<isconst>
,即當前實例。)
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