问题
I have a class like
template <class T>
struct A{
template <class U>
A(U u);
};
I would like to write an explicit specialization of this for a declaration like
A<int>::A(float);
In the following test code, if I comment out the specialization, it compiles with g++. Otherwise, it says I have the wrong number of template parameters:
#include <iostream>
template <class T>
struct A{
template <class U>
A(T t, U *u){
*u += U(t);
}
};
template <>
template <>
A<int>::A<int,float>(int t, float *u){
*u += float(2*t);
}
int main(){
float f = 0;
int i = 1;
A<int>(i, &f);
std::cout << f << std::endl;
return 0;
}
回答1:
Try
template <>
template <>
A<int>::A(int t, float *u){
*u += float(2*t);
}
That seems to work for me.
回答2:
The definition's function parameter list should match the declaration's.
template <>
template <>
A<int>::A<float>(int t, float *u){
*u += U(2*t);
}
来源:https://stackoverflow.com/questions/2563165/c-explicit-template-specialization-of-templated-constructor-of-templated-class