CODEWITHSUNDEEP
c++templatesc++11tuplestype-traits
Vincent
Vincent

Reputation: 60381

Detect if a type is a std::tuple?

Currently I have two functions :

template<typename Type> bool f(Type* x);
template<typename... List> bool f(std::tuple<List...>* x);

Is there any way to merge these two functions with an extra template parameter that indicates whether the passed type is a tuple ?

template<typename Type, bool IsTuple = /* SOMETHING */> bool f(Type* x);

Upvotes: 21

Views: 12145

Answers (5)

smac89
smac89

Reputation: 43088

With C++17, here is a fairly simple solution using if constexpr

template <typename> struct is_tuple: std::false_type {};

template <typename ...T> struct is_tuple<std::tuple<T...>>: std::true_type {};

Then you can do something like:

template<typename Type> bool f(Type* x) {
    if constexpr (is_tuple<Type>::value) {
        std::cout << "A tuple!!\n";
        return true;
    }

    std::cout << "Not a tuple\n";
    return false;
}

A test to ensure it worked:

f(&some_tuple);
f(&some_object);

Output:

A tuple!!
Not a tuple

Solution taken in part from an answer found here: How to know if a type is a specialization of std::vector?

Upvotes: 18

David Ledger
David Ledger

Reputation: 2101

Might be a little late but you can also do something like this, in a more modern c++17 style with template variables:

template <typename T>
constexpr bool IsTuple = false;
template<typename ... types>
constexpr bool IsTuple<std::tuple<types...>>   = true;

And some tests

struct TestStruct{};

static_assert(IsTuple<int> == false,                "Doesn't work with literal.");
static_assert(IsTuple<TestStruct> == false,         "Doesn't work with classes.");
static_assert(IsTuple<std::tuple<int, char>>,       "Doesn't work with plain tuple.");
static_assert(IsTuple<std::tuple<int&, char&>>,     "Doesn't work with lvalue references");
static_assert(IsTuple<std::tuple<int&&, char&&>>,   "Doesn't work with rvalue references");

You can view it here https://godbolt.org/z/FYI1jS

EDIT: You will want to run std::decay, std::remove_volatile, std::remove_const to handle special cases.

Upvotes: 4

Dave Dopson
Dave Dopson

Reputation: 42694

With C++11, this is my preferred pattern:

// IsTuple<T>()
template <typename T>
struct IsTupleImpl : std::false_type {};

template <typename... U>
struct IsTupleImpl<std::tuple <U...>> : std::true_type {};

template <typename T>
constexpr bool IsTuple() {
  return IsTupleImpl<decay_t<T>>::value;
}

Works great. No dependencies (I can't use Boost).

Upvotes: 6

Xeo
Xeo

Reputation: 131789

Sure, using is_specialization_of (link taken and fixed from here):

template<typename Type, bool IsTuple = is_specialization_of<Type, std::tuple>::value>
bool f(Type* x);

The question is, however, do you really want that? Normally, if you need to know if a type is a tuple, you need special handling for tuples, and that usually has to do with its template arguments. As such, you might want to stick to your overloaded version.

Edit: Since you mentioned you only need a small portion specialized, I recommend overloading but only for the small special part:

template<class T>
bool f(T* x){
  // common parts...
  f_special_part(x);
  // common parts...
}

with

template<class T>
void f_special_part(T* x){ /* general case */ }

template<class... Args>
void f_special_part(std::tuple<Args...>* x){ /* special tuple case */ }

Upvotes: 19

Vaughn Cato
Vaughn Cato

Reputation: 64308

You could just have your functions defer to another function:

template<typename Type,bool IsTuple> bool f(Type *x);

template<typename Type> 
inline bool f(Type* x) { return f<Type,false>(x); }

template<typename... List> 
inline bool f(std::tuple<List...>* x) { return f<std::tuple<List...>,true>(x); }

Upvotes: 5

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