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// Fallback, anything with an operator() template <typename T> struct function_traits : public function_traits<decltype(&T::operator())> > { };
What does T::operator() mean here?
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2 Answers
Reset to default
8
&T::operator() is a pointer to member function. In this case it’s a pointer to typename T‘s operator() member function.
To give this a concrete grounding, consider this minimal definition of T:
struct T
{
int operator()(int x) { return x * 2; }
};
We could save a pointer to the member function in a variable:
auto pmf = &T::operator();
And we can call it:
T object;
int four = (object.*pmf)(2);
0
The function call operator allows an object to be called as a function.
e.g.
#include <iostream>
#include <string>
class Howdy {
std::string name_;
public:
Howdy(const std::string& name) : name_(name) {}
void operator()() const { // <- this is "operator()"
std::cout << "Hello " << name_ << std::endl;
}
};
int main() {
Howdy howdy{"world"};
howdy(); // Outputs "Hello world"
}
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2
This doesn't really explain what the expression
&T::operator()does, what type it creates, and how it is relevant to traits.– Jan Schultke14 hours ago
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