Function handle in varargin not recognized

Question

Mikael Freyhult 2022-6-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1742040-function-handle-in-varargin-not-recognized

编辑： Voss 2022-7-3

When i am trying to input a function manually the scripts doesent recognize x ( Unrecognized function or variable 'x'. ) see pictures

How am i using the function handle wrong?

Thanks

function drawfunction(varargin)
switch nargin
    case 0
        disp('0 inparam. chosen')
        f = @(x) exp(atan(x)+sin(x))+log(2+x.^4)-(cos(3*x)/(1+x.^2));
        x = linspace(-3,8);
        color = 'r:';
        
    case 1
        disp('1 inparam. chosen')
        f = @(x) varargin(1);
        x = linspace(-3,8);
        color = 'r:';
        
    case 2
        disp('2 inparam. chosen')
        f = @(x) varargin(1);
        x = varargin(2);
        color = 'r:';
        
    case 3
        disp('3 inparam. chosen')
        f = @(x)varargin(1);
        x = varargin(2);
        color = varargin(3);
        
    otherwise
        error('Unexpexted inputs')
end
plot(f(x),color)

No input:

One input: (same function)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Voss 2022-6-16

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1742040-function-handle-in-varargin-not-recognized#answer_987245

在 MATLAB Online 中打开

You're calling drawfunction with an expression that MATLAB tries to evaluate before sending the result of that evaluation in to drawfunction.

What you mean to do is to send a function handle to drawfunction, so call it like this:

drawfunction( @(x)exp(atan(x)+sin(x)) ) % ( etc., long expression )

Then, inside drawfunction, varargin{1} (note the curly braces, by the way) is that function handle you passed in, so there's no need to do this:

f = @(x)varargin(1) % neither one of these is correct
f = @(x)varargin{1}

It's just this:

f = varargin{1}; % varargin{1}, i.e., the first input argument, is already a function handle

3 个评论
显示 1更早的评论隐藏 1更早的评论

Voss 2022-6-16

编辑：Voss 2022-6-16

在 MATLAB Online 中打开

Another way to specify a function as an input argument is as a character vector, e.g.:

func = 'exp(atan(x)+sin(x))';

If you want drawfunction to handle arguments like that, you can convert them to function handles inside drawfunction using str2func:

f = str2func(['@(x)' varargin{1}]); % assuming the function is a function of x

You can handle both char-vector and function-handle inputs by checking the class of varargin{1} and converting it if necessary:

if ischar(varargin{1}) % convert character vector to function_handle
    f = str2func(['@(x)' varargin{1}]); % assuming the function is a function of x
else % already a function_handle
    f = varargin{1};
end

And, by the way, now that that's a few lines of code, so you don't have to copy and paste it several times for the case when nargin is 2, 3, 4, etc., you can modify your argument handling structure a little bit:

function drawfunction(varargin)
if nargin >= 1 % at least one argument given
    if ischar(varargin{1})
        f = str2func(['@(x)' varargin{1}]);
    else
        f = varargin{1};
    end
else
    f = @(x)exp(atan(x)+sin(x)); % default function
end
if nargin >= 2 % at least two arguments given
    x = varargin{2};
else
    x = linspace(-3,8); % default x
end
if nargin >= 3 % at least three arguments given
    color = varargin{3};
else
    color = 'r:'; % default color/line-spec
end
if nargin >= 4
%     error('Unexpexted inputs');
    warning('drawfunction: Extra arguments ignored.');
else
    disp(sprintf('%d inparam. chosen',nargin));
end
plot(x,f(x),color);
end

Mikael Freyhult 2022-6-16

Thank you very much!

I understand I will need to "create the function handle" when calling the function, or as you described use srting2func. I was a bit confused as why it didn't just pass the argument to varagin{1} but then i understand that it evaluates it "before" sending it into the function.

Sorry if my terminology is under par :P

If i may ask a subsequent question, is there any way to, without changing the code to enter a blank argument as varagin{1} and varagin{2} then inputing varagin{3}?

I didn't manage it by leaving blanks as -> drawfunction( , ,'b:')

to only change the plot, see image

i tried increasing number of cases to include only for example varagin{3} but it did not work

%%% Code below

function drawfunction(varargin)

switch nargin

case 0

disp('0 inparam. chosen')

f = @(x) exp(atan(x)+sin(x))+log(2+x.^4)-(cos(3*x)/(1+x.^2));

x = linspace(-3,8);

color = 'r:';

case 1

disp('first inparam. chosen')

f = varargin{1};

x = linspace(-3,8);

color = 'r:';

case 2

disp('1,2 inparam. chosen')

f = varargin{1};

x = varargin{2};

color = 'r:';

case 3

disp('1,2,3 inparam. chosen')

f = varargin{1};

x = varargin{2};

color = varargin{3};

case 4

disp('Third inparam. chosen')

f = @(x) exp(atan(x)+sin(x))+log(2+x.^4)-(cos(3*x)/(1+x.^2));

x = linspace(-3,8);

color = varargin{3};

case 5

disp('2,3 inparam. chosen')

f = @(x) exp(atan(x)+sin(x))+log(2+x.^4)-(cos(3*x)/(1+x.^2));

x = varargin{2};

color = varargin{3};

case 6

disp('1,3 inparam. chosen')

f = varargin{1};

x = linspace(-3,8);

color = varargin{3};

otherwise

error('Unexpexted inputs')

end

plot(f(x),color)

Voss 2022-6-16

编辑：Voss 2022-6-16

在 MATLAB Online 中打开

As you've discovered, you cannot send missing arguments like that

drawfunction( , , 'b:')

If you want to use an input argument, then all previous input arguments must also be specified.

However, you can write your function such that it handles certain values specially. For instance, you might use the empty array [] as a special value that indicates to the function that the default value for that argument should be used (this is only useful if [] is not a valid value for that argument).

If you do that, then you could call drawfunction like this:

drawfunction([], [], 'b:')

3 inparam. chosen

to get the default function and x values. Of course, you also have to set up drawfunction to handle [] appropriately, which is easy if you use the argument handling structure I showed in my last comment:

function drawfunction(varargin)
if nargin >= 1 && ~isempty(varargin{1}) % at least one argument given and it's not empty
    if ischar(varargin{1})
        f = str2func(['@(x)' varargin{1}]);
    else
        f = varargin{1};
    end
else
    f = @(x)exp(atan(x)+sin(x)); % default function
end
if nargin >= 2 && ~isempty(varargin{2}) % at least two arguments given and argument 2 is not empty
    x = varargin{2};
else
    x = linspace(-3,8); % default x
end
if nargin >= 3 && ~isempty(varargin{3}) % at least three arguments given and argument 3 is not empty
    color = varargin{3};
else
    color = 'r:'; % default color/line-spec
end
if nargin >= 4
%     error('Unexpexted inputs');
    warning('drawfunction: Extra arguments ignored.');
else
    disp(sprintf('%d inparam. chosen',nargin));
end
plot(x,f(x),color);
end

(Note that I changed

plot(f(x),color);

to

plot(x,f(x),color);

because I think that makes more sense.)

请先登录，再进行评论。

Function handle in varargin not recognized

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

3 个评论
显示 1更早的评论隐藏 1更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Function handle in varargin not recognized

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（1 个）

3 个评论 显示 1更早的评论隐藏 1更早的评论

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

3 个评论
显示 1更早的评论隐藏 1更早的评论