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I am trying to write a template class which may or may not define a particular member function depending on its template parameter type. Further the return type of this member function depends on the return type of of a member of the template paramter (if defined).

Below is a minimal example of my code

#include <iostream>
#include <type_traits>

template <typename T>
struct has_foo_int {
private:
    template <typename U>
    static decltype(std::declval<U>().foo(0), void(), std::true_type()) test(int);
    template <typename>
    static std::false_type test(...);
public:
    typedef decltype(test<T>(0)) test_type;
    enum { value = test_type::value };
};

template <typename T, bool HasFooInt>
struct foo_int_return_type;

template<typename T>
struct foo_int_return_type<T,false> {};

template<typename T>
struct foo_int_return_type<T,true> {
    using type = decltype(std::declval<T>().foo(0));
};

template<typename T>
struct mystruct
{
    T val;

    //auto someMethod(int i) -> decltype(std::declval<T>().foo(0)) // error: request for member ‘foo’ in ‘std::declval<double>()’, which is of non-class type ‘double’
    //auto someMethod(int i) -> typename foo_int_return_type<T,has_foo_int<T>::value>::type // error: no type named ‘type’ in ‘struct foo_int_return_type<double, false>’
    template<typename R=typename foo_int_return_type<T,has_foo_int<T>::value>::type> R someMethod(int i) // error: no type named ‘type’ in ‘struct foo_int_return_type<double, false>’
    {
        return val.foo(i);
    }
};

struct with_foo_int {
    int foo(int i){
        return i+1;
    }
};

using namespace std;

int main(void)
{
    mystruct<with_foo_int> ms1;
    cout << ms1.someMethod(41) << endl;

    mystruct<double> ms2;

    return 0;
}

What I would like to happen is that the code compiles fine and outputs 42 for ms1.someFunc(41). I would also expect that if one accidentally tried to call someFunc on ms2 that it would fail to compile.

Unfortunately each of the alternatives I have tried has failed. The first and second, I think I understand why they wouldn't work.

I read here that SFINAE only works for template functions so I tried giving a dummy template parameter to work out the return type but this too fails in the same way.

I'm clearly not understanding something here, what am I missing? Is it possible to achieve what I'm trying to do?

Thanks.

P.s. I'm using g++ 4.7.3

P.p.s I have also tried std::enable_if but get much the same results as with my foo_int_return_type struct.

4

1 回答 1

2

这是一种简短、整洁且记录在案的方法来做你正在尝试的事情，之后解决了一些可能的错误。

#include <type_traits>

/*
    Template `has_mf_foo_accepts_int_returns_int<T>`
    has a static boolean public member `value` that == true
    if and only if `T` is a class type that has a public
    member function or member function overload 
    `int T::foo(ArgType) [const]`  where `ArgType`
    is a type to which `int` is implicitly convertible.
*/
template <typename T>
struct has_mf_foo_accepts_int_returns_int {

    /* SFINAE success:
        We know now here `int *` is convertible to
        "pointer to return-type of T::foo(0)" 
    */
    template<typename A>
    static constexpr bool test(
        decltype(std::declval<A>().foo(0)) *prt) {
        /* Yes, but is the return-type of `T::foo(0)`
            actually *the same* as `int`?...
        */
        return std::is_same<int *,decltype(prt)>::value;
    }

    // SFINAE failure :(
    template <typename A>
    static constexpr bool test(...) {
        return false; 
    }

    /* SFINAE probe.
        Can we convert `(int *)nullptr to 
        "pointer to the return type of T::foo(0)"?
    */
    static const bool value = test<T>(static_cast<int *>(nullptr)); 
};

template<typename T>
struct mystruct
{
    using has_good_foo = has_mf_foo_accepts_int_returns_int<T>;

    T val;

    /*  SFINAE:
        `template<typename R> R someMethod(R)` will be this if and only
        if `R` == `int` and `has_good_foo` == true.         
    */
    template<typename R = int>
    typename std::enable_if<
        (has_good_foo::value && std::is_same<R,int>::value),R
    >::type 
    someMethod(R i) {
        return val.foo(i);
    }

    /*  SFINAE:
        `template<typename R> R someMethod(R)` will be this if and only
        if `R` != `int` or `has_good_foo` != true.      
    */
    template<typename R = int>
    typename std::enable_if<
        !(has_good_foo::value && std::is_same<R,int>::value),R
    >::type
    someMethod(R i) {
        static_assert(has_good_foo::value && std::is_same<R,int>::value,
            "mystruct<T> does not implement someMethod(R)");
        return i;
    }
};

// Testing...

#include <iostream>

struct with_foo_int
{
    int foo(int i) {
        return i + 1;
    }
};

using namespace std;

int main(void)
{
    mystruct<with_foo_int> ms1;
    cout << ms1.someMethod(41) << endl;

    mystruct<double> ms2;
    cout << ms2.someMethod(41) << endl; // static_assert failure

    return 0;
}

该解决方案忠实地重现了您自己尝试中的几个可能的漏洞，如下所示：-

1)看起来您可能认为评估std::declval<U>().foo(0)是一种确定是否U::foo存在的 SFINAE 方法，并采用 type 的单个参数int。它没有。它只是一种 SFINAE 方法，用于确定是否 U::foo(ArgType)存在ArgType任何可0隐式转换的东西。因此ArgType可以是任何指针或算术类型，而不仅仅是int.

2）您可能没有考虑过如果其中一个或两个存在std::declval<U>().foo(0)就会满足。您可能很关心您是否在 上调用一个或非成员函数 ，并且您肯定会关心您调用了两个成员函数中的哪一个。如果 定义为：U::foo(ArgType) U::foo(ArgType) constconstconstUwith_foo_int

struct with_foo_int
{
    int foo(int i) const {
        return i + 1;
    }
    int foo(int i) {
        return i + 2;
    }   
};

那么给出的解决方案将调用非const重载 ms1.someMethod(41)并将 == 43。

2）容易处理。如果您希望确保只能调用， T::foo(ArgType) const请const在mystruct::someMethod. 如果您不关心或只想打电话T::foo(ArgType)，请保持原样。

1) 解决起来有点困难，因为您必须制作一个 SNIFAE 探针， T::foo只有当它具有正确的签名时才会满足该探针，并且该签名将是const合格的或不合格的。让我们假设你想要 int T::foo(int) const. 在这种情况下，将模板替换 has_mf_foo_accepts_int_returns_int为：

/*  Template `has_mf_foo_arg_int_returns_int<T>
    has a static boolean public member `value` that == true
    if and only if `T` is a class type that has an un-overloaded
    a public member `int T::foo(int) const`.
*/ 
template< typename T>
struct has_mf_foo_arg_int_returns_int
{
    /* SFINAE foo-has-correct-sig :) */
    template<typename A>
    static std::true_type test(int (A::*)(int) const) {
        return std::true_type();
    }

    /* SFINAE foo-exists :) */
    template <typename A> 
    static decltype(test(&A::foo)) 
    test(decltype(&A::foo),void *) {
        /* foo exists. What about sig? */
        typedef decltype(test(&A::foo)) return_type; 
        return return_type();
    }

    /* SFINAE game over :( */
    template<typename A>
    static std::false_type test(...) {
        return std::false_type(); 
    }

    /* This will be either `std::true_type` or `std::false_type` */
    typedef decltype(test<T>(0,0)) type;

    static const bool value = type::value; /* Which is it? */
};

并在模板中mystruct替换：

using has_good_foo = has_mf_foo_accepts_int_returns_int<T>;

和：

using has_good_foo = has_mf_foo_arg_int_returns_int<T>;

（模板has_mf_foo_arg_int_returns_int改编自我的其他答案，您可以在那里阅读它的工作原理。）

从后一种方法中获得的 SFINAE 精度是有代价的。该方法要求您尝试获取 的地址T::foo，以查看它是否存在。但是 C++ 不会给你一个重载的成员函数的地址，所以如果T::foo重载，这个方法就会失败。

static_assert此处的代码将使用 GCC >= 4.7.2 clang >= 3.2编译（或适当地）。

于 2013-07-30T09:11:15.620 回答