[英]“expected primary expression” error on template method using
我有一些實現帕累托規則的通用代碼。 它似乎是格式良好的代碼。
關於newResult.set<Criterion>( criterion() );
錯誤的GCC 4.4編譯器消息newResult.set<Criterion>( criterion() );
表達。 但我找不到問題。
完整錯誤日志:
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
完整代碼清單:
// 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);
}
找到了:
newResult.template set<Criterion>( criterion() );
if(newResult.template get<Criterion>() >= oldResult.template get<Criterion>())
return false;
在這種情況下,您必須限定編譯器的成員函數模板。 詞法分析器將無法決定(在模板聲明時 ,不是實例化時 ) <Criterion
是指模板參數列表的開始,還是比較運算符。
看到
標准,§14.2,sub 4.和5. ,值得注意的:
[注意:與typename前綴的情況一樣,在不是絕對必要的情況下允許使用模板前綴; 即,當嵌套名稱說明符或 - >或者左邊的表達式時。 不依賴於模板參數,或者使用不會出現在模板的范圍內。 - 尾注]
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