代码之家 › 专栏 › 技术社区 › Tony The Lion

运算符=模棱两可(C++)

ambiguous operators boost c++

Tony The Lion · 技术社区 · 14 年前

我在for循环中有以下内容,编译器说“operator=不明确”。不知道怎么解决这个问题,有人能帮忙吗?

rootelement = document->getDocumentElement();
    boost::interprocess::unique_ptr<DOMNodeIterator, release_deleter> itera (document->createNodeIterator(rootelement, DOMNodeFilter::SHOW_ALL, NULL, true));
    for(boost::interprocess::unique_ptr<DOMNode, release_deleter> current (itera->nextNode()); current != 0; current = boost::interprocess::unique_ptr<DOMNode, release_deleter> (itera->nextNode()))  // Last assignment on current is ambiguous

完全错误:

\XMLDocument.cpp(193) : error C2593: 'operator =' is ambiguous
        c:\Program Files\boost\boost_1_44\boost\interprocess\smart_ptr\unique_ptr.hpp(249): could be 'boost::interprocess::unique_ptr<T,D> &boost::interprocess::unique_ptr<T,D>::operator =(int boost::interprocess::unique_ptr<T,D>::nat::* )'
        with
        [
            T=xercesc_3_1::DOMNode,
            D=release_deleter
        ]
        unique_ptr.hpp(211): or       'boost::interprocess::unique_ptr<T,D> &boost::interprocess::unique_ptr<T,D>::operator =(boost::interprocess::rv<boost::interprocess::unique_ptr<T,D>> &)'
        with
        [
            T=xercesc_3_1::DOMNode,
            D=release_deleter
        ]
        while trying to match the argument list '(boost::interprocess::unique_ptr<T,D>, boost::interprocess::unique_ptr<T,D>)'
        with
        [
            T=xercesc_3_1::DOMNode,
            D=release_deleter
        ]
        and
        [
            T=xercesc_3_1::DOMNode,
            D=release_deleter
        ]
  \XMLDocument.cpp(193) : see reference to class template instantiation 'boost::interprocess::detail::unique_ptr_error<T>' being compiled
        with
        [
            T=boost::interprocess::unique_ptr<xercesc_3_1::DOMNode,release_deleter>
        ]
    XMLDocument.cpp(193) : see reference to class template instantiation 'boost::interprocess::detail::unique_ptr_error<T>' being compiled
        with
        [
            T=xercesc_3_1::DOMNode *
        ]
       \XMLDocument.cpp(193) : see reference to class template instantiation 'boost::interprocess::detail::unique_ptr_error<T>' being compiled
        with
        [
            T=xercesc_3_1::DOMNode *
        ]
        XMLDocument.cpp(192) : see reference to class template instantiation 'boost::int
erprocess::detail::unique_ptr_error<T>' being compiled
        with
        [
            T=xercesc_3_1::DOMNodeIterator *
        ]

3 回复 | 直到 14 年前

GManNickG 14 年前

就像斯蒂芬说的, unique_ptr 除非您显式地移动它们,或者为它们分配一个右值,否则它们保持唯一性。通常情况下,您的代码会很好,但是由于您在伪造rvalues,所以需要显式地移动它。

我从没用过 boost::interprocess::unique_ptr ,但看起来您需要:

namespace bi = boost::interprocess; // do these please!
typedef bi::unique_ptr<DOMNode, release_deleter> node_ptr;
typedef bi::unique_ptr<DOMNodeIterator, release_deleter> iterator_ptr;

rootelement = document->getDocumentElement();
iterator_ptr itera(document->createNodeIterator(rootelement,
                                           DOMNodeFilter::SHOW_ALL, NULL, true));

for (node_ptr current(itera->nextNode()); current != 0;
         current = bi::move(node_ptr(itera->nextNode())))

更简单的可能是:

for (node_ptr current(itera->nextNode()); current != 0;
         current.reset(itera->nextNode()))

Stephane Rolland 14 年前

我认为STD::UnQuyJPTR只能被分配给STD:: 这就是它明确地失去底层对象所有权的方式。

std ::unique_ptr<T> upOldT = new T ;
std ::unique_ptr<T> pT = std ::move(upOldT) ;

正如GMan所说的,C++0X还不是当前的C++标准,所以应该是 boost::进程间::唯一指针…

Jan 14 年前

我不确定,也没有尝试过任何东西,但您可以尝试:

current = itera->nextNode()

那就只需要其中一个模棱两可了。