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SFINAE Constructors [duplicate]

Tags:

c++

sfinae

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) {
}
like image 825
Evan Teran Avatar asked Jul 11 '18 19:07

Evan Teran


1 Answers

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.

like image 130
Justin Avatar answered Oct 10 '22 23:10

Justin