I saw C++ lambda source.
#include <functional>
#include <iostream>
int main()
{
std::function<int(int)> factorial;
// factorial = [factorial](int n)->int // runtime error
factorial = [&factorial](int n)->int // right
{
if (n == 1) {
return 1;
}
else {
return n * factorial(n - 1);
}
};
std::cout << factorial(5) << "\n";
}
I don't understand why runtime error occurs.
Thank you for your concerning!
factorial = [factorial](int n)->int
This version captures factorial by value. At the point of the lambda expression factorial is empty, so you get a copy of an empty std::function and you'll get a std::bad_function_call if you try to call it. The factorial = /*...*/; is assigning to the original object, which is separate from the copy created for the closure.
factorial = [&factorial](int n)->int
This version captures factorial by reference, i.e. factorial inside the lambda names the same object as factorial outside. As such, the assignment factorial = /*...*/; affects the factorial variable inside the lambda, so you call a valid function.
