Hump-day puzzler - Unknown Function

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My colleague was working with MATLAB and defined an anonymous function F. When he saw me come in he typed clc, hit return than challenged me to guess the form of the function only by calling it however many times I want. After calling it many times, I have figured out that it takes a scalar argument, and when the value of the argument is greater than 0 it does this:
>> F(1)
Name Size Bytes Class Attributes
---- 0;']) ------------------------------------------
x 1x1 8 double
And when the value of the argument is less than zero, it appears to do nothing:
>> F(-1)
Can you help me figure out what the function looks like?
Matt Fig
Matt Fig on 25 May 2011
Hint #2. Here's how I know the function wants a scalar argument...
>> = 't';
>> F(S)
>> F({3})
>> F(magic(2))

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Accepted Answer

Laura Proctor
Laura Proctor on 25 May 2011
I get the sense that this is not exactly what the user did, but this seems to do what is expected as a first try:
F=@(x) eval('if isscalar(x)&&x>0, whos x, end')
Second iterate: I needed to add in a check for type double. It also prints out the funky characters with WHOS (but without any functionality). :oP
F=@(x) eval(['if isscalar(x)&&isa(x,''double'')&&x>0, whos x, end; 1>0;'])
Matt Fig
Matt Fig on 25 May 2011
@Oleg, you aren't the only one who doesn't understand why this happens. I was hoping Laura could shed some light. See my previous comment.

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More Answers (1)

Sean de Wolski
Sean de Wolski on 25 May 2011
Tough one! In lieu of a real way to name a function/function handle: 0;'])
F = @(x)HDC525(x);
function HDC525(x)
%SCd 5/25/2011: Lame attempt at HD challenger
if ~isscalar(x)||x<=0
Str = evalc('whos x');
fprintf('%s\n ---- 0;'']) ------------------------------------------\n %s',Str(1:numel(Str)/2-1),Str(numel(Str)/2+1:end));
Sean de Wolski
Sean de Wolski on 25 May 2011
I know it doesn't. And I know that 'whos' will put the function handle's meat where you have _0;'])_ but I know how to overwrite it. Arghhhhh.

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