C++ Templated class function that can detect std::vector - c++17

Question

If I have a templated class, I can do the following to detect if a vector was passed:
template<typename T> struct is_vector { static const bool value=false; };
template<typename T> struct is_vector<std::vector<T>> { static const bool value=true; };
template<class T>
class Parser {
public:
Parser() {}
void parse(T obj) {
if (is_vector<T>::value) {
std::cout << "vector\n";
//obj.push_back(T {});
}
else {
std::cout << "not vector\n";
}
}
};
int main() {
Parser<int> p1;
p1.parse(123);
Parser<std::vector<int>> p2;
p2.parse({ 1, 2, 3});
return 0;
}
Output:
not vector
vector
I can detect a vector, yet the compiler complains when I uncomment the push_back call:
main.cpp: In instantiation of ‘void Parser<T>::parse(T) [with T = int]’:
main.cpp:26:14: required from here
main.cpp:15:17: error: request for member ‘push_back’ in ‘obj’, which is of non-class type ‘int’
obj.push_back(T {});
~~~~^~~~~~~~~
Obviously, an int does not have a push_back function, but the vector does. The is_vector call is evaluated at runtime, but the push_back is caught at compile time.
With partial template specialization, I can do what I want:
template<typename T>
void parse(T obj) {
std::cout << "not vector: " << obj << "\n";
}
template<typename T>
void parse(std::vector<T> obj) {
std::cout << "is vector\n";
for (auto i : obj) std::cout << i << " ";
obj.push_back(T {});
std::cout << "\n";
for (auto i : obj) std::cout << i << " ";
std::cout << "\n";
}
int main() {
parse(1);
parse('a');
parse(std::vector<int> { 1, 2, 3 });
return 0;
}
Output:
not vector: 1
not vector: a
is vector
1 2 3
1 2 3 0
So, how can I combine these 2 ideas, either at compile-time or at runtime? That is, have a templated class with a function that can handle vectors and non-vectors?

Answer 1

What you're looking for is a new feature in C++17, if constexpr. It's the same as a regular if, except that the condition is evaluated at compile time, and when instantiating the branch(es) will discard the non-taken branch at compile time. The discarded branch does not need to well-formed. So, for your example:
template<class T>
class Parser {
public:
Parser() {}
void parse(T obj) {
if constexpr (is_vector<T>::value) {
std::cout << "vector\n";
obj.push_back(T {});
}
else {
std::cout << "not vector\n";
}
}
};
See Difference between if constexpr vs if for some more talk on the differences. You can also read the cppreference page on if statements to get a detailed overview of some of the nitty-gritty details.