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我从具有部分专业化的模板继承，并且不能从派生的 ctor 调用模板 ctor。

当下面代码中的部分特化被注释掉时，它编译时不会出现任何错误或警告。

#include <iostream>
typedef enum {supertype, subtype} selector;

template< typename T, selector s>
class Tmpl {
protected:
    T* root;
public:
    Tmpl( T* t = 0 ) {
        root = t;
    }
    T listHead( ) {
        std::cout << "template listHead() called" << std::endl;
    }
};

class Descriptor {
public:
    Descriptor( const char * s ) {
        std::cout << "Descriptor " << s << std::endl;
    }
};

// partial specialization - if uncommented, errors 
// are reported at the supertypesIterator ctor below.
/*
template<selector s>
class Tmpl<Descriptor, s> {
public:
    Descriptor listHead( ) {
        switch( s ){
            case supertype:
                return Descriptor("Supertypes");
            case subtype:
                return Descriptor("Subtypes");
        }
    }
};
*/

class supertypesIterator : public Tmpl<Descriptor, supertype> {
public:
    supertypesIterator( Descriptor* t = 0 ):Tmpl<Descriptor, supertype>(t) {}
};


main() {
    supertypesIterator s;
    s.listHead();
}

如果我取消注释专业化，我会收到以下错误：

$g++ trouble.cc

trouble.cc: In constructor ‘supertypesIterator::supertypesIterator(Descriptor*)’:
trouble.cc:43:74: error: no matching function for call to ‘Tmpl<Descriptor, (selector)0u>::Tmpl(Descriptor*&)’
trouble.cc:43:74: note: candidates are:
trouble.cc:27:7: note: Tmpl<Descriptor, (selector)0u>::Tmpl()
trouble.cc:27:7: note:   candidate expects 0 arguments, 1 provided
trouble.cc:27:7: note: Tmpl<Descriptor, (selector)0u>::Tmpl(const Tmpl<Descriptor, (selector)0u>&)
trouble.cc:27:7: note:   no known conversion for argument 1 from ‘Descriptor*’ to ‘const Tmpl<Descriptor, (selector)0u>&’

我需要做什么才能从supertypesIteratorctor 中初始化基类？

我正在使用 g++ version 4.7.1，尽管我还需要它才能跨平台工作。

4

1 回答 1

2

您必须在您的专业化中实现缺少的构造函数。否则，构造函数supertypesIterator会尝试调用Tmpl不存在的构造函数。

template<selector s>
class Tmpl<Descriptor, s> {
    Descriptor* root;
public:
    Tmpl( Descriptor* t = 0 ) {
        root = t;
    }
    Descriptor listHead( ) {
        switch( s ){
            case supertype:
                return Descriptor("Supertypes");
            case subtype:
                return Descriptor("Subtypes");
        }
    }
};
于 2012-07-30T22:57:37.023 回答