Matlab functions handles and variable and unknown number of outputs

Question

With an anonymous function, you can return any number of outputs. What I need is to be able to use functors (anonymous functions as arguments of other functions), while not knowing how many outputs I will get.

This is to avoid code duplication by injecting functions calls inside a while loop which is reused in many functions.

Example:

function y = foo( x )  
    y = x;
end

function [y1, y2] = goo( x1, x2 )  
    y1 = x1;
    y2 = x2;
end

function [ varargout ] = yolo( functor, varargin )  
    varargout = functor(varargin{:});    
end

I want to be able to call:

y = yolo(@foo, 2)
[y1, y2] = yolo(@goo, 3, 4);

Is there any way to achieve this ? Thanks

Suever · Accepted Answer · 2016-11-15 16:00:44Z

4

It is not possible to get the number of outputs of an anonymous function (a function handle to an inline function) because the output is always varargout and therefore nargout is always going to return -1

myfunc = @(x, y) x + y;
nargout(myfunc)
%   -1

However, it looks like you don't have anonymous functions, but rather just function handles to normal functions that are defined in an .m file and have an explicit number of output arguments. In this case, you can combine nargout with {:} indexing to fill varargout with all of the output arguments.

function y = foo(x)
    y = x;
end

function [y1, y2] = goo(x1, x2)
    y1 = x1;
    y2 = x2;
end

function varargout = yolo(functor, varargin)
    [varargout{1:nargout(functor)}] = functor(varargin{:});
end

y = yolo(@foo, 2)
[y1, y2] = yolo(@goo, 3, 4)

edited Nov 15, 2016 at 16:00

answered Nov 15, 2016 at 15:41

Suever

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1 Comment

Milan Over a year ago

Thanks for your answer, I was trying to look for a workaround using the number of outputs as an argument but it seems this will work, going to check this right now. I did not know I could use nargout as a function

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