C++ SFINAE 示例？

Question

I want to get into more template meta-programming. I know that SFINAE stands for "substitution failure is not an error." But can someone show me a good use for SFINAE?

Answer 1

I like using SFINAE to check boolean conditions.

template<int I> void div(char(*)[I % 2 == 0] = 0) {
    /* this is taken when I is even */
}

template<int I> void div(char(*)[I % 2 == 1] = 0) {
    /* this is taken when I is odd */
}

It can be quite useful. For example, i used it to check whether an initializer list collected using operator comma is no longer than a fixed size

template<int N>
struct Vector {
    template<int M> 
    Vector(MyInitList<M> const& i, char(*)[M <= N] = 0) { /* ... */ }
}

The list is only accepted when M is smaller than N, which means that the initializer list has not too many elements.

The syntax char(*)[C] means: Pointer to an array with element type char and size C . If C is false (0 here), then we get the invalid type char(*)[0] , pointer to a zero sized array: SFINAE makes it so that the template will be ignored then.

Expressed with boost::enable_if , that looks like this

template<int N>
struct Vector {
    template<int M> 
    Vector(MyInitList<M> const& i, 
           typename enable_if_c<(M <= N)>::type* = 0) { /* ... */ }
}

In practice, i often find the ability to check conditions a useful ability.

Answer 2

Heres one example ( from here ):

template<typename T>
class IsClassT {
  private:
    typedef char One;
    typedef struct { char a[2]; } Two;
    template<typename C> static One test(int C::*);
    // Will be chosen if T is anything except a class.
    template<typename C> static Two test(...);
  public:
    enum { Yes = sizeof(IsClassT<T>::test<T>(0)) == 1 };
    enum { No = !Yes };
};

When IsClassT<int>::Yes is evaluated, 0 cannot be converted to int int::* because int is not a class, so it can't have a member pointer. If SFINAE didn't exist, then you would get a compiler error, something like '0 cannot be converted to member pointer for non-class type int'. Instead, it just uses the ... form which returns Two, and thus evaluates to false, int is not a class type.

Answer 3

In C++11 SFINAE tests have become much prettier. Here are a few examples of common uses:

Pick a function overload depending on traits

template<typename T>
std::enable_if_t<std::is_integral<T>::value> f(T t){
    //integral version
}
template<typename T>
std::enable_if_t<std::is_floating_point<T>::value> f(T t){
    //floating point version
}

Using a so called type sink idiom you can do pretty arbitrary tests on a type like checking if it has a member and if that member is of a certain type

//this goes in some header so you can use it everywhere
template<typename T>
struct TypeSink{
    using Type = void;
};
template<typename T>
using TypeSinkT = typename TypeSink<T>::Type;

//use case
template<typename T, typename=void>
struct HasBarOfTypeInt : std::false_type{};
template<typename T>
struct HasBarOfTypeInt<T, TypeSinkT<decltype(std::declval<T&>().*(&T::bar))>> :
    std::is_same<typename std::decay<decltype(std::declval<T&>().*(&T::bar))>::type,int>{};


struct S{
   int bar;
};
struct K{

};

template<typename T, typename = TypeSinkT<decltype(&T::bar)>>
void print(T){
    std::cout << "has bar" << std::endl;
}
void print(...){
    std::cout << "no bar" << std::endl;
}

int main(){
    print(S{});
    print(K{});
    std::cout << "bar is int: " << HasBarOfTypeInt<S>::value << std::endl;
}

Here is a live example: http://ideone.com/dHhyHE I also recently wrote a whole section on SFINAE and tag dispatch in my blog (shameless plug but relevant) http://metaporky.blogspot.de/2014/08/part-7-static-dispatch-function.html

Note as of C++14 there is a std::void_t which is essentially the same as my TypeSink here.

Answer 4

Boost's enable_if library offers a nice clean interface for using SFINAE. One of my favorite usage examples is in the Boost.Iterator library. SFINAE is used to enable iterator type conversions.

Answer 5

C++17 will probably provide a generic means to query for features. See N4502 for details, but as a self-contained example consider the following.

This part is the constant part, put it in a header.

// See http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4502.pdf.
template <typename...>
using void_t = void;

// Primary template handles all types not supporting the operation.
template <typename, template <typename> class, typename = void_t<>>
struct detect : std::false_type {};

// Specialization recognizes/validates only types supporting the archetype.
template <typename T, template <typename> class Op>
struct detect<T, Op, void_t<Op<T>>> : std::true_type {};

The following example, taken from N4502 , shows the usage:

// Archetypal expression for assignment operation.
template <typename T>
using assign_t = decltype(std::declval<T&>() = std::declval<T const &>())

// Trait corresponding to that archetype.
template <typename T>
using is_assignable = detect<T, assign_t>;

Compared to the other implementations, this one is fairly simple: a reduced set of tools ( void_t and detect ) suffices. Besides, it was reported (see N4502 ) that it is measurably more efficient (compile-time and compiler memory consumption) than previous approaches.

Here is a live example , which includes portability tweaks for GCC pre 5.1.

Answer 6

Here's another (late) SFINAE example, based on Greg Rogers 's answer :

template<typename T>
class IsClassT {
    template<typename C> static bool test(int C::*) {return true;}
    template<typename C> static bool test(...) {return false;}
public:
    static bool value;
};

template<typename T>
bool IsClassT<T>::value=IsClassT<T>::test<T>(0);

In this way, you can check the value 's value to see whether T is a class or not:

int main(void) {
    std::cout << IsClassT<std::string>::value << std::endl; // true
    std::cout << IsClassT<int>::value << std::endl;         // false
    return 0;
}

Answer 7

Here is one good article of SFINAE: An introduction to C++'s SFINAE concept: compile-time introspection of a class member .

Summary it as following:

/*
 The compiler will try this overload since it's less generic than the variadic.
 T will be replace by int which gives us void f(const int& t, int::iterator* b = nullptr);
 int doesn't have an iterator sub-type, but the compiler doesn't throw a bunch of errors.
 It simply tries the next overload. 
*/
template <typename T> void f(const T& t, typename T::iterator* it = nullptr) { }

// The sink-hole.
void f(...) { }

f(1); // Calls void f(...) { }

---

template<bool B, class T = void> // Default template version.
struct enable_if {}; // This struct doesn't define "type" and the substitution will fail if you try to access it.

template<class T> // A specialisation used if the expression is true. 
struct enable_if<true, T> { typedef T type; }; // This struct do have a "type" and won't fail on access.

template <class T> typename enable_if<hasSerialize<T>::value, std::string>::type serialize(const T& obj)
{
    return obj.serialize();
}

template <class T> typename enable_if<!hasSerialize<T>::value, std::string>::type serialize(const T& obj)
{
    return to_string(obj);
}

---

declval is an utility that gives you a "fake reference" to an object of a type that couldn't be easily construct. declval is really handy for our SFINAE constructions.

struct Default {
    int foo() const {return 1;}
};

struct NonDefault {
    NonDefault(const NonDefault&) {}
    int foo() const {return 1;}
};

int main()
{
    decltype(Default().foo()) n1 = 1; // int n1
//  decltype(NonDefault().foo()) n2 = n1; // error: no default constructor
    decltype(std::declval<NonDefault>().foo()) n2 = n1; // int n2
    std::cout << "n2 = " << n2 << '\n';
}

Answer 8

The following code uses SFINAE to let compiler select an overload based on whether a type has certain method or not:

    #include <iostream>
    
    template<typename T>
    void do_something(const T& value, decltype(value.get_int()) = 0) {
        std::cout << "Int: " <<  value.get_int() << std::endl;
    }
    
    template<typename T>
    void do_something(const T& value, decltype(value.get_float()) = 0) {
        std::cout << "Float: " << value.get_float() << std::endl;
    }
    
    
    struct FloatItem {
        float get_float() const {
            return 1.0f;
        }
    };
    
    struct IntItem {
        int get_int() const {
            return -1;
        }
    };
    
    struct UniversalItem : public IntItem, public FloatItem {};
    
    int main() {
        do_something(FloatItem{});
        do_something(IntItem{});
        // the following fails because template substitution
        // leads to ambiguity 
        // do_something(UniversalItem{});
        return 0;
    }

Output:

Float: 1
Int: -1

Answer 9

A new blog exists since the last answer to this thread.

It is Fluent C++: http://fluentcpp.com/

There are plenty of examples for a research on "SFINAE".

Answer 10

Here, I am using template function overloading (not directly SFINAE) to determine whether a pointer is a function or member class pointer: ( Is possible to fix the iostream cout/cerr member function pointers being printed as 1 or true? )

https://godbolt.org/z/c2NmzR

#include<iostream>

template<typename Return, typename... Args>
constexpr bool is_function_pointer(Return(*pointer)(Args...)) {
    return true;
}

template<typename Return, typename ClassType, typename... Args>
constexpr bool is_function_pointer(Return(ClassType::*pointer)(Args...)) {
    return true;
}

template<typename... Args>
constexpr bool is_function_pointer(Args...) {
    return false;
}

struct test_debugger { void var() {} };
void fun_void_void(){};
void fun_void_double(double d){};
double fun_double_double(double d){return d;}

int main(void) {
    int* var;

    std::cout << std::boolalpha;
    std::cout << "0. " << is_function_pointer(var) << std::endl;
    std::cout << "1. " << is_function_pointer(fun_void_void) << std::endl;
    std::cout << "2. " << is_function_pointer(fun_void_double) << std::endl;
    std::cout << "3. " << is_function_pointer(fun_double_double) << std::endl;
    std::cout << "4. " << is_function_pointer(&test_debugger::var) << std::endl;
    return 0;
}

Prints

0. false
1. true
2. true
3. true
4. true

As the code is, it could (depending on the compiler "good" will) generate a run time call to a function which will return true or false. If you would like to force the is_function_pointer(var) to evaluate at compile type (no function calls performed at run time), you can use the constexpr variable trick:

constexpr bool ispointer = is_function_pointer(var);
std::cout << "ispointer " << ispointer << std::endl;

By the C++ standard, all constexpr variables are guaranteed to be evaluated at compile time ( Computing length of a C string at compile time. Is this really a constexpr? ).

Answer 11

Examples provided by other answers seems to me more complicated than needed.

Here is the slightly easier to understand example from cppreference :

#include <iostream>
 
// this overload is always in the set of overloads
// ellipsis parameter has the lowest ranking for overload resolution
void test(...)
{
    std::cout << "Catch-all overload called\n";
}
 
// this overload is added to the set of overloads if
// C is a reference-to-class type and F is a pointer to member function of C
template <class C, class F>
auto test(C c, F f) -> decltype((void)(c.*f)(), void())
{
    std::cout << "Reference overload called\n";
}
 
// this overload is added to the set of overloads if
// C is a pointer-to-class type and F is a pointer to member function of C
template <class C, class F>
auto test(C c, F f) -> decltype((void)((c->*f)()), void())
{
    std::cout << "Pointer overload called\n";
}
 
struct X { void f() {} };
 
int main(){
  X x;
  test( x, &X::f);
  test(&x, &X::f);
  test(42, 1337);
}

Output:

Reference overload called
Pointer overload called
Catch-all overload called

As you can see, in the third call of test, substitution fail without errors.