I can use enable_if to separate behavior by parameter type such as:
std::vector<int>
Now I want to separate behavior by the inner type of a container:
int of std::vector<int>
what can I do in c++?
I wonder if this is what you mean:
#include<iostream>
#include<vector>
#include<type_traits>
// The following function will display the contents of the provided T
// only if its value_type trait is of type int.
template<typename T>
typename std::enable_if<
std::is_same<typename T::value_type, int>::value,
void>::type display(const T& data) {
std::cout<<"Some ints:"<<std::endl;
for(int xi : data) {
std::cout<<xi<<" ";
}
std::cout<<std::endl;
}
int main() {
std::vector<int> dint = {1, 2, 3, 4, 5, 6};
display(dint);
std::vector<float> dfloat = {1, 2, 3, 4, 5, 6};
// The following call would lead to compile-time error, because
// there is no display function activated for float types:
//display(dfloat);
return 0;
}
Compiling with g++ example.cpp -std=c++11 -Wall -Wextra
(OS X 10.7.4 using GCC 4.8.1) yields:
$ ./a.out
Some ints:
1 2 3 4 5 6
As expected, if I uncomment the display(dfloat)
line the compiler error message includes:
error: no matching function for call to ‘display(std::vector<float>&)’