问题
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) {
}
回答1:
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.
回答2:
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.
来源：https://stackoverflow.com/questions/51292574/sfinae-constructors