nargin of optional arugments

function func(arg1,options)
    arguments
        arg1 = 1
        options.arg2 = 2
        options.arg3 = 3
    end
    disp(nargin)
end

%% 
func(1,arg2=3,arg3=5)

     1

The function "func" has 3 input arguments, why nargin is 1? And how to get the number of all input arguments, including optional ones?

[Noé]

Hi Wang,

As mentioned in https://mathworks.com/help/releases/R2024a/matlab/matlab_prog/nargin-in-argument-validation.html "The value that nargin returns does not include optional input arguments that are not included in the function call. Also, nargin does not count any name-value arguments."

Here arg2 and arg3 are optional arguments whereas arg1 is positional argument.

If you want to know how many arguments have been passed you can:

"Use nargin to determine if optional positional arguments are passed to the function when called. For example, this function declares three positional arguments and a name-value argument. Here is how the function determines what arguments are passed when it is called.

• nargin determines if the optional positional argument c is passed to the function with a switch block.
• isfield determines if the name-value argument for Format is passed to the function."

[Catalytic]

Also, nargin does not count any name-value arguments

I find this a bit of an unsettling design choice. Why not create a new command with this behaviour rather than change the behaviour of old ones?

Answer 1

[Catalytic]

func(1,arg2=3,arg3=5)

NUM_ARGSIN = 3

options = struct with fields:
    arg2: 3
    arg3: 5

function func(arg1,names,values)
    arguments
        arg1 = 1
    end
    arguments (Repeating)
        names  string
        values double
    end
    
    NUM_ARGSIN = numel(names)+1
    
    nvp=[names;values];
    options=struct(nvp{:})
end

[Matt J]

If you're going to look at it that way, I think it should be,

func()

NUM_ARGSIN = 0

func(10)

NUM_ARGSIN = 1

func(1,arg2=3,arg3=5)

NUM_ARGSIN = 3

function func(arg1,names,values)
    arguments
        arg1 = 1
    end
    arguments (Repeating)
        names  string
        values double
    end
    
    NUM_ARGSIN = nargin - numel(names)
    
    nvp=[names;values];
    options=struct(nvp{:});
end

[Hupeng]

Thank you.

Answer 2

[Matt J]

The function "func" has 3 input arguments, why nargin is 1?

There are 5 input arguments, not 3:

func(1,arg2=3,arg3=5)

NUM_ARGSIN = 5

arg1 = 1

options = struct with fields:
    arg2: 3
    arg3: 5

function func(varargin)

    NUM_ARGSIN = nargin

    [arg1,options]=doValidations(varargin{:});
    disp ' ', arg1, options
    
end


function [arg1,options]=doValidations(arg1,options)
    arguments
        arg1 = 1
        options.arg2 = 2
        options.arg3 = 3
    end
end