检查函数签名也继承功能 [英] Check for function signature also for inherited functions
问题描述
我需要检查，如果一个货柜擦除函数返回迭代器。我通常检查通过例如函数签名促进。但是，在升压级的情况下（例如flat_set）擦除继承，因此不会被检查发现。但我真的需要它。 <一href=\"https://stackoverflow.com/questions/1966362/sfinae-to-check-for-inherited-member-functions\">SFINAE检查继承成员函数只能说明我还不能使用C ++ 11的解决方案。
我想是这样的：
模板＆LT; typename的T＆GT;
类has_member_func_erase_it_constit
{
的typedef typename的T ::迭代器迭代器;
的typedef typename的T ::为const_iterator为const_iterator;
的typedef BOOST_TYPEOF_TPL（＆amp; T公司::擦除）eraseType; 的typedef typename的提振:: function_types :: result_type的＆LT; eraseType＆GT; ::类型的搜索结果;
上市：
静态常量布尔值=提振:: is_same＆LT;迭代器，导致＆GT; ::值;
}; 模板＆LT;类T＆GT;
结构EraseReturnsIterator
{
静态CONSTEXPR布尔值= has_member_func_erase_it_constit＆LT; T＆GT; ::值;
};
但由于擦除过载失败。我可能会需要decltype或类似的东西，以检查与常量性抹去编译时调用的返回类型，但我不能找到一个。
这怎么可能在pre C ++ 11？
如果有一个擦除功能返回void这也并不工作：
模板＆LT; typename的T＆GT;
类has_member_func_erase_it
{
的typedef typename的T ::迭代器迭代器;
的typedef typename的T ::为const_iterator为const_iterator; 的typedef CHAR是[1];
焦炭的typedef无[2]; 静态ŧMAKET（）;
静态迭代makeIt（）; 的typedef BOOST_TYPEOF_TPL（MAKET（）擦除（makeIt（）））的结果; 上市：
静态常量布尔值=提振:: is_same＆LT;迭代器，导致＆GT; ::值;
};
以下工作：
///房屋类型
模板＆LT; typename的T＆GT;结构Type2Type {
typedef的T形;
}; ///邪恶两轮牛车把一个前pression成式
模板＆LT; typename的T＆GT;
Type2Type＆LT; T＆GT;运营商（T，Type2Type＆LT;无效＆GT;）; 模板＆LT; typename的T＆GT;
ŧdeclval（）; 模板＆LT;类T＆GT;
结构EraseReturnsIterator
{
的typedef typename的T ::迭代器迭代器;
的typedef BOOST_TYPEOF_TPL（（declval＆LT; T＆GT;（）擦除（declval＆LT;＆迭代器GT;（）），＆Type2Type LT;无效＆GT;（）））的结果; 静态CONSTEXPR布尔值=提振:: is_same＆LT;迭代器，类型名称结果::类型＆GT; ::值;
};
Basicly我们只需要调用的返回类型，我们需要的功能。如果没有这种类型的函数返回永远比无效就BOOST_TYPEOF_TPL已经工作。不幸的是擦除可以返回无效，因为它试图通过打破实行无效＆安培;某处堆栈。
因此，为了避免虚空（没有双关语意），我们把它放在一个类型，即持有型。为了能够为前pressions做到这一点，我们重载逗号操作符。这样的结果等于Type2Type，我们可以轻松地阅读。完成！
理念为逗号过载： http://stackoverflow.com/a/5553460/1930508
I need to check, if a containers erase function returns the iterator. I'd normally check for the function signature via e.g. boost. But in case of a boost class (e.g. flat_set) erase is inherited and thus not found by the check. But I really need it. SFINAE to check for inherited member functions only shows a C++11 solution which I can't use yet.
I tried something like this:
template <typename T>
class has_member_func_erase_it_constit
{
typedef typename T::iterator iterator;
typedef typename T::const_iterator const_iterator;
typedef BOOST_TYPEOF_TPL(&T::erase) eraseType;
typedef typename boost::function_types::result_type<eraseType>::type result;
public:
static const bool value = boost::is_same<iterator, result>::value;
};
template<class T>
struct EraseReturnsIterator
{
static CONSTEXPR bool value = has_member_func_erase_it_constit<T>::value;
};
But it fails since erase is overloaded. I'd probably need decltype or something like that to check the return type from a compile-time invocation of erase with const_iterator, but I can't find one.
How is this possible in pre C++11?
This does also not work if there is an erase function returning void:
template <typename T>
class has_member_func_erase_it
{
typedef typename T::iterator iterator;
typedef typename T::const_iterator const_iterator;
typedef char yes[1];
typedef char no [2];
static T makeT();
static iterator makeIt();
typedef BOOST_TYPEOF_TPL(makeT().erase(makeIt())) result;
public:
static const bool value = boost::is_same<iterator, result>::value;
};
The following works:
/// "Stores a type"
template<typename T> struct Type2Type{
typedef T type;
};
/// Evil hackery to put an expression into a type
template<typename T>
Type2Type<T> operator,(T, Type2Type<void>);
template<typename T>
T declval();
template<class T>
struct EraseReturnsIterator
{
typedef typename T::iterator iterator;
typedef BOOST_TYPEOF_TPL((declval<T>().erase(declval<iterator>()), Type2Type<void>())) result;
static CONSTEXPR bool value = boost::is_same<iterator, typename result::type>::value;
};
Basicly we just call the function whose return type we need. If no function of this type ever returned void than BOOST_TYPEOF_TPL would already work. Unfortunately erase CAN return void which breaks the implementation as it tries to pass "void&" somewhere down the stack.
So to avoid the void (no pun intended) we put it in a type, that holds a type. To be able to do this for expressions, we overload the comma operator. This way result equals "Type2Type" which we can easily read. Done!
Idea for the comma-overload: http://stackoverflow.com/a/5553460/1930508
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