Assume we have class that depends on two template types one of we specialize in constructor, can we not specialize deducable type?
template <typename T, typename I>
class A {
public:
A(I i) {
}
};
int main() {
A<float, int> a(42); // this works, but we can deduce int from 42
A<float> aa(42); // doesn't work
//or
auto aaa = A<float>(42); // doesn't work
}
In C++20, you can use alias template deduction like this:
template <typename I>
using A_float = A<float, I>;
int main() {
A_float a(42);
auto aa = A_float(42);
}
Try it on godbolt.org: Demo.
Alternatively, you can define a make function and deduce I from the function parameter. Because of copy elision, this is no less efficient than constructing the A directly:
template <typename T, typename I>
auto make_a(I i) { return A<T, I>(i); }
int main() {
auto a = make_a<float>(42);
}
Try it on godbolt.org: Demo.
