Flow of information in callback functions

I am using a 3rd party optimization solver which comes with examples and I am trying to understand it. An example in it uses callback functions in these formats.

% The callback functions.
funcs.f1        = @(x) x(1)*x(4)*sum(x(1:3)) + x(3);
funcs.f2          = @f2;
funcs.f3 = @() sparse(ones(2,4));
   
where 

function g = f2 (x)
g=f(x)% g is some function of x
end

And it is run as 
 [x] = solver(x0,funcs);
where x is the output vector of variables.

I am having trouble understanding the difference between flow of information in the 3 formats of the callback functions in the funcs struct.

1. In the funcs.f1, @(x) tells that it is a function with x as variables, right?

2. What does @f2 in funcs.f2 mean? Is the output g of the function f2 being passed in this format to funcs.f2?

3. And does @() in funcs.f3 mean that it is not the function of variable x and hence the input argument to @() is blank?

I would appreciate if someone would help.

Answer 1

[Rik]

You can find some more information in the documentation.

The @ creates a function handle. There are two options here: an anonymous function, or a normal function handle.

The f1 statement creates an anonymous function with a single input. Any time you call it, it will run as if the x is replaced by your input variable, and all other variables are the same as before.

n=1;
f1=@(x) x+n;
n=2;
v=-1;
f1(v)%returns 0 because it uses the old value of n

The f2 statement creates a function handle. When you run it, Matlab will look for that function on the search path. You can use it in functions like accumarray where you need to input a function handle (or you could even use it to shorten function names).

f2=@mean;
f2([1 2 3])%returns 2, same as mean([1 2 3]) would

The f3 statement creates an anonymous function without an input. Like any function this can have any number of output arguments.

f3=@() disp('next round..')
for n=1:3
    trigger_boxing_bell_function
    f3();
end

Hope this helps.

[DB]

Thanks, it clarifies a lot.

[Rik]

Your welcome. If my answer helped you, please consider marking it as accepted answer.