Question

我有一些实现帕累托规则的通用代码。它似乎是格式良好的代码。

GCC 4.4关于newResult.set<Criterion>( criterion() );表达式错误的编译器消息。但我找不到问题。

完整错误日志：

trunk$ g++ -std=c++0x -o test test.cpp 
t6.cpp: In member function ‘bool Pareto<Minimize<T>, Types ...>::operator()(Map&, Map&)’:
t6.cpp:24: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘)’ token
t6.cpp:26: error: expected primary-expression before ‘>’ token
t6.cpp:26: error: expected primary-expression before ‘)’ token
t6.cpp: In member function ‘bool Pareto<Maximize<T>, Types ...>::operator()(Map&, Map&)’:
t6.cpp:43: error: expected primary-expression before ‘>’ token
t6.cpp:45: error: expected primary-expression before ‘>’ token
t6.cpp:45: error: expected primary-expression before ‘)’ token
t6.cpp:45: error: expected primary-expression before ‘>’ token
t6.cpp:45: error: expected primary-expression before ‘)’ token

完整代码列表：

// TypeMap
template < typename ... Tail >
struct Holder;

template <typename ValueType, typename Head, typename ... Tail >
struct Holder<ValueType, Head, Tail ... > :
    public Holder<ValueType, Head>,
    public Holder<ValueType, Tail ... >
{};

template <typename ValueType, typename Head >
struct Holder<ValueType, Head>
{
    ValueType value;
};

template < typename ... Types >
struct TypeMap;

template <typename ValueType, typename ... Types >
struct TypeMap<ValueType, Types ... > :
    public Holder<ValueType, Types ... >
{
    template <typename T>
    void set(const ValueType& value)
    {
        ((Holder<ValueType, T>*)this)->value = value;
    }

    template <typename T>
    ValueType get()
    {
        return ((Holder<ValueType, T>*)this)->value;
    }
};

// Objectives
template <typename Criterion> struct Maximize : public Criterion {};
template <typename Criterion> struct Minimize : public Criterion {};

// Criteria
struct Criterion1{ double operator()(){ return 0; }};
struct Criterion2{ double operator()(){ return 0; }};

// Pareto rule
template < typename ... Types > struct Pareto;

template < typename T, typename ... Types >
struct Pareto<Minimize<T>, Types ... >
{
    template< typename Map >
    bool operator()(Map& oldResult, Map& newResult)
    {
        typedef Minimize<T> Criterion;
        Criterion criterion;

        // ERROR HERE !!!
        newResult.set<Criterion>( criterion() );

        if(newResult.get<Criterion>() >= oldResult.get<Criterion>())
            return false;

        Pareto<Types ... > pareto;
        return pareto(oldResult, newResult);
    }
};

template < typename T, typename ... Types >
struct Pareto<Maximize<T>, Types ... >
{
    template< typename Map >
    bool operator()(Map& oldResult, Map& newResult)
    {
        typedef Maximize<T> Criterion;
        Criterion criterion;

        // ERROR HERE !!!
        newResult.set<Criterion>( criterion() );

        if(newResult.get<Criterion>() <= oldResult.get<Criterion>())
            return false;

        Pareto<Types ... > pareto;
        return pareto(oldResult, newResult);
    }
};

template<>
struct Pareto<>
{
    template<typename Map>
    bool operator()(Map& oldResult, Map& newResult)
    {
        oldResult = newResult;
        return true;
    }
};

int main()
{
    TypeMap<double, Minimize<Criterion1>, Maximize<Criterion2>> oldResult, newResult;

    Pareto<Minimize<Criterion1>, Maximize<Criterion2>> pareto;
    pareto(oldResult, newResult);
}

Answer 1

找到它：

newResult.template set<Criterion>( criterion() );

if(newResult.template get<Criterion>() >= oldResult.template get<Criterion>())
    return false;

在这种情况下，您必须限定编译器的成员函数模板。词法分析器无法决定（在模板声明时，而不是实例化），<Criterion是否意味着模板参数的开始列表或者比较运算符。

见

Using the template keyword as qualifier
What is the .template and ::template syntax about?（Comeau）
标准，§14.2，sub 4。和 5。，值得注意的是：

[注意：与typename前缀的情况一样，如果是，则允许使用模板前缀不是绝对必要的;即，当嵌套名称说明符或左侧的表达式 - >要么。不是依赖于模板参数，或者使用不会出现在模板的范围内。 - 后注]

使用模板方法的“预期主表达式”错误

1 个答案: