Distinguish uifigure from figure programmatically

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Adam on 12 Jul 2017
Commented: Adam Danz on 21 May 2021
Is it possible to distinguish a uifigure (appdesigner) from an old-style figure programmatically by an if type test?
class( hFig )
for both of them and I can't see anything obvious in the metaclass or properties that I can use to check which it is.
I could just use a try-catch, but I'd prefer to just filter uifigures than catch the error (I have a BusyCursor class which changes the pointer on all open figures to the busy cursor and then back again when it is deleted, but it crashes on uifigures which do not support changes to the 'pointer' property).
Adam Danz
Adam Danz on 21 May 2021
It depends on what you're doing with the figure. Try/catch won't do squat if you need to work with the figure position property, for example, since the two types of figures have different position definitions and getting/setting position won't cause an error.

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Accepted Answer

Ralph Coleman
Ralph Coleman on 18 Sep 2019
Edited: Ralph Coleman on 18 Sep 2019
From R2018b, you can simply use:
Adam Danz
Adam Danz on 21 Apr 2021
A combination of the top 2 solutions here that cover all existing matlab releases to test if a figure handle h is a uifigure.
if verLessThan('Matlab','9.0') %version < 16a (release of uifigs)
isuifig = @(~)false;
elseif verLessThan('Matlab','9.5') % 16a <= version < 18b
isuifig = @(h)~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
else % version >= 18b (written in r21a)
isuifig = @(h)matlab.ui.internal.isUIFigure(h);
tf = isuifig(h);
...however regular figures generated in live editor are incorrectly identified as UIFigures using Ralf's method 🤨. I haven't tested it thoroughly but a potential workaround is to test for the presence of an undocumented property of figures generated in live editor.
isuifig = @(h)matlab.ui.internal.isUIFigure(h) && ~isprop(h,'LiveEditorRunTimeFigure');

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More Answers (4)

Zhengyi on 18 Feb 2019
Edited: Zhengyi on 18 Feb 2019
function val = isuifigure(h)
val = ~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
The slightly longer story
If you type the following into the console, you will get an error:
>> uialert(figure,'a','b')
Error using uialert (line 42)
First argument must be a figure handle created using the uifigure function.
On line 42 of uialert function, we can see it uses the following method to validate the input handle:
By reading through this function, we can see the error is thrown when the following function returns empty:
function out = getFigureID(f)
out = f.getId();
seems to be a private function. The only option we have is to use getFigureID and check whether it returns a non-empty value.
Finaly, if you like, you can create this utility function to help you get over the long function call:
function val = isuifigure(h)
val = ~isempty(matlab.ui.internal.dialog.DialogHelper.getFigureID(h));
  1 Comment
Adam Danz
Adam Danz on 4 Mar 2021
Thanks for digging, @Zhengyi. This solution works all the way back to r2016a which is when uifigures were released. However, is breaks in r2020b (and, perhaps, prior releases) when the handle was created by figure().
% Error:
% First argument must be a figure handle created using the uifigure function.
A workaround would be to use a try/catch and to test the MException for the error ID 'MATLAB:uitools:uidialogs:NotAnAppWindowHandle' but, according to your answer, that line returns an empty value in earlier releases and who knows if this error message will change in the future.

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Dev-iL on 15 Aug 2017
I tried comparing the outputs of:
F = struct(uifigure());
G = struct(figure());
There are several fields that are different by default. Of increased interest are:
JavaFrame, Controller, ControllerInfo, NodeChildren
I cannot guarantee that it will work 100% of the time, but you can try testing for
... if the above returns "true" - it's likely a uifigure.

Andreas Klotzek
Andreas Klotzek on 9 Jan 2019
I also need an answer to this question.
The answers provided here are not certain to give the correct result. For example if I have a old-style figure with property HandleVisibility = 'off', the test using groot will give the wrong result.
Using properties is also not a stable idea. The behavior might change.
The problem is not only for figure and uifigure. I need the same check on axes/uiaxes or uipanel handles, basically on any graphics handle. One could probably use the ancestor function to redirect the problem for any graphics handle to the figure handle.

Celso Reyes
Celso Reyes on 20 Aug 2018
Edited: Celso Reyes on 21 Aug 2018
The new figures do not show in groot by default -- At least as of Mac version, R2018a -- when ShowHiddenHandles is 'off'. If ShowHiddenHandles is 'on', then it will appear.
As a side note, the new-style uifigure appears when you use allchild(groot) regardless of the ShowHiddenHandles value.
So, one could non-invasive check would look like so...
function style = figtype(f)
g = groot;
oldStatus = g.ShowHiddenHandles;
g.ShowHiddenHandles = 'off';
if any(g.Children == f)
style = 'figure';
style = 'uifigure';
g.ShowHiddenHandles = oldStatus;
So, in practice...
>> newf = uifigure('Name','new');
>> oldf = figure('Name','old');
>> figtype(newf)
ans =
>> figtype(oldf)
ans =
I'm not sure what would make a regular figure's handle hidden, since the Visibility property doesn't do it.

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