问题
Here is my question: I define a functor:
class A {
public:
int operator()(int a, int b) const{
return a + b;
}
};
typedef function<int (int, int)> Fun;
then I use a anonymous functor to create a std::function object, and I find something strange. Here is my code:
Fun f(A());
f(3, 4);
Unfortunately it is wrong. The error message is:
error: invalid conversion from ‘int’ to ‘A (*)()’ [-fpermissive]
error: too many arguments to function ‘Fun f(A (*)())’
However, when I change my code as follow:
A a;
Fun f(a);
f(3, 4);
or
Fun f = A();
f(3, 4);
The result is right. So, why is it? Help me understand it,please. Thanks.
回答1:
Fun f(A());
This is a case of the most-vexing parse. It declares a function f
which returns a Fun
. It takes a function pointer as an argument, pointing at a function that takes no arguments and returns an A
.
There are a few ways to get around this:
Fun f{A()}; // Uniform-initialisation syntax
Fun f{A{}}; // Uniform-initialisation on both objects
Fun f((A())); // Forcing the initialiser to be an expression, not parameter list
Or one of the things you did.
来源:https://stackoverflow.com/questions/27252065/pass-anonymous-function-object-to-stdfunction