SFINAE Constructors [duplicate]

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SFINAE working in return type but not as template parameter 3 answers
Template specialization and enable_if problems [duplicate] 1 answer

I have been liking SFINAE syntax like this for functions, seems to generally work well!

template <class Integer, class = typename std::enable_if<std::is_integral<Integer>::value>::type>
T(Integer n) {
    // ...
}

But am running into an issue when I want to do this as well in the same class...

template <class Float, class = typename std::enable_if<std::is_floating_point<Float>::value>::type>
T(Float n) {
    // ...
}

Getting errors such as this:

./../T.h:286:2: error: constructor cannot be redeclared
        T(Float n) {
        ^
./../T.h:281:2: note: previous definition is here
        T(Integer n) {
        ^
1 error generated.

Shouldn't these constructors only exist for the appropriate types and never at the same time? Why do they conflict?

Am I being a bit thick here?

This on the other hand does work (but I don't like the syntax as much):

template <class Integer>
T(Integer n, typename std::enable_if<std::is_integral<Integer>::value>::type* = nullptr) {
}

template <class Float>
T(Float n, typename std::enable_if<std::is_floating_point<Float>::value>::type* = nullptr) {
}

Use a non-type template parameter instead:

template <class Integer,
    std::enable_if_t<std::is_integral<Integer>::value, int> = 0>
T(Integer n) {
    // ...
}

template <class Float,
    std::enable_if_t<std::is_floating_point<Float>::value, int> = 0>
T(Float n) {
    // ...
}

This works because the compiler has to substitute the first template parameter before it can determine the type of the value parameter.

One way to fix this is to add extra , typename=void arguments so that none of the overloads has the same number of template parameters.