I found this type trait which can be used to check if a certain type T supports operator<<:
template<class Class>
struct has_ostream_operator_impl {
template<class V>
static auto test(V*) -> decltype(std::declval<std::ostream>() << std::declval<V>());
template<typename>
static auto test(...) -> std::false_type;
using type = typename std::is_same<std::ostream&, decltype(test<Class>(0))>::type;
};
template<class Class>
struct has_ostream_operator : has_ostream_operator_impl<Class>::type {};
Source: https://gist.github.com/szymek156/9b1b90fe474277be4641e9ef4666f472
This works fine. Now I'm trying to do the same thing for operator>> using c++11, but I don't get it to work:
template<class Class>
struct has_istream_operator_impl {
template<class V>
static auto test(V*) -> decltype(std::declval<V>() >> std::declval<std::istream>());
template<typename>
static auto test(...) -> std::false_type;
using type = typename std::is_same<std::istream&, decltype(test<Class>(0))>::type;
};
/**
* @brief Type trait to check if operator>>(std::istream, Type) is defined for a given type.
*/
template<class Class>
struct has_istream_operator : has_istream_operator_impl<Class>::type {};
Here is a simplified test for my use case:
#include <sstream>
#include <type_traits>
#include <iostream>
// <include snippet 2>
template<typename T>
typename std::enable_if<has_istream_operator<T>::value, T>::type
fromString(const std::string& str) {
T value;
std::istringstream stream(str);
stream >> value;
return value;
}
int main() {
std::cout << fromString<long>("123") + 1 << std::endl; // expecting 124
return 0;
}
Error is:
has_istream_operator.cpp: In function ‘int main()’:
has_istream_operator.cpp:57:38: error: no matching function for call to ‘fromString<long int>(const char [4])’
std::cout << fromString<long>("123") + 1 << std::endl; // expecting 124
^
has_istream_operator.cpp:49:1: note: candidate: ‘template<class T> typename std::enable_if<has_istream_operator<Class>::value, T>::type fromString(const string&)’
fromString(const std::string& str) {
^~~~~~~~~~
has_istream_operator.cpp:49:1: note: template argument deduction/substitution failed:
has_istream_operator.cpp: In substitution of ‘template<class T> typename std::enable_if<has_istream_operator<Class>::value, T>::type fromString(const string&) [with T = long int]’:
has_istream_operator.cpp:57:38: required from here
has_istream_operator.cpp:49:1: error: no type named ‘type’ in ‘struct std::enable_if<false, long int>’
From which I understand that the SFINAE condition is false and therefore no definition of fromString exists.
What I have tried is to play around with the line
static auto test(V*) -> decltype(std::declval<V>() >> std::declval<std::istream>());
inside my definition of struct has_istream_operator_impl.
This is the variation which makes most sense to me, because when I use the operator>>, I usually do it this way: stream >> value for example and from my (limited) understanding, the test(V*) should test this generically for V:
static auto test(V*) -> decltype(std::declval<std::istream>() >> std::declval<V>());
But it does also not work (same error).
How do get I this to work?
Long story short, you should change
static auto test(V*) -> decltype(std::declval<V>() >> std::declval<std::istream>());
to
static auto test(V*) -> decltype(std::declval<std::istream>() >> std::declval<V&>());
There were two errors in the code, due to the following.
>> operator takes the stream as first argument, not as second argument, whereas you are passing the two arguments the other way around.declval<V>() is generating an rvalue (well, not really a value because we are in an unevaluated context), to which you can't assign a value via >> (just like you can't cin >> 123;), so you have to change it to declval<V&>().(¹)In reality, prompted by @brahmin's comment, I have to point out another bug in the original code. Follow this reasoning:
has_istream_operator_impl<T>::test is correctly chosen if >> makes sense between an std::istream (any value category) and a T& (lvalue);has_istream_operator_impl<T>::type will incorrectly test if the return type of the call to test<T>(0) is std::istream&.operator>> and they don't all return std::istream&:
static_assert(std::is_same_v<
decltype(std::declval<std::istream>() >> std::declval<std::string&>()),
std::istream&&>);
static_assert(std::is_same_v<
decltype(std::declval<std::istream>() >> std::declval<int&>()),
std::istream&>);
What you want to do, is to have has_istream_operator_impl<T>::type be std::true_type or std::false_type only depending on whether std::declval<std::istream>() >> std::declval<V&>() is valid , not on depending on its return type!
One typical approach to do so is to use decltype with two arguments (or to static_cast<void> the result of one-argument decltype)
static auto test(V*)
-> decltype(std::declval<std::istream>() >> std::declval<V&>(), void());
using type = typename std::is_same<void, decltype(test<Class>(0))>::type;
So the correct code would be the following
template<typename T>
struct has_istream_operator_impl {
template<typename V>
static auto test(void*) -> decltype(std::declval<std::istream>() >> std::declval<V&>(), void());
template<typename>
static auto test(...) -> std::false_type;
static constexpr bool value = std::is_same_v<decltype(test<T>(nullptr)), void>;
};
template<typename T>
constexpr bool has_istream_operator = has_istream_operator_impl<T>::value;
where I've made some cosmetic changes:
has_istream_operator a value rather than a type, so the usage is less verbose:
static_assert(has_istream_operator<int>);
static_assert(!has_istream_operator<int*>);
static_assert(has_istream_operator<std::string>);
void* rather than V* for the "positve" test function parameter, as that has nothing to do with with V,0 to nullptr because we are passing it to a function taking a pointer,class to typename as using both is confusing or at least distracting, imho.¹ To understand more in depth why that's the case, look at the possible implementation of std::declval as shown at the documentation page on cppreference: as you can see, it returns the type typename std::add_rvalue_reference<T>::type (which, incidentally, can be written as std::add_rvalue_reference_t<T> since C++14), i.e. std::declval<T>() returns T&&, which is (by reference collapsing)
T to std::declval (e.g. std::declval<int&>()),std::declval<int>()).In your usecase you are passing long as the T to std::declval, so we are in the second case, i.e. std::declval<long>() returns long&&. From the page on value categories you can see that an example of xvalue (which is, just like a prvalue, an rvalue) is the following (my emphasis):
a function call or an overloaded operator expression, whose return type is rvalue reference to object, such as std::move(x);
And that's exactly what std::declval<long>() is: it has a rvalue reference to object as its return type, hence it returns an rvalue.
If you call, instead, std::declval<T&>() and pass long as T, the call will be std::declval<long&>(), in which case the return type will be long&, hence the call will return an lvalue. See the following quote, from the same page, of an example of lvalue
a function call or an overloaded operator expression, whose return type is lvalue reference;
To give an example of what std::declval is doing, these both pass
static_assert(std::is_same_v<decltype(std::declval<int>()), int&&>);
static_assert(std::is_same_v<decltype(std::declval<int&>()), int&>);
and this fails to compile
int x;
std::cin >> static_cast<int&&>(x); // XXX
where the // XXX line is what std::declval<std::istream>() >> std::declval<V>() is "emulating".