Hump-day puzzler - Unknown Function

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?

2 个评论

Hint #1. The function does not call an M-File
Hint #2. Here's how I know the function wants a scalar argument...
>> S.one = 't';
>> F(S)
>> F({3})
>> F(magic(2))
>>

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 采纳的回答

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;'])

5 个评论

You are on the right track! Nice work...
Your second iteration is close enough! Good MATLAB detective work, Laura
I don't understand what's happening!!!
Do you know why WHOS prints this way when used like that? I.E., why it behaves differently than it does here:
F=@(x) eval(['if isscalar(x)&&isa(x,''double'')&&x<0, whos x, end; 1<0;'])
@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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更多回答(1 个)

Tough one! In lieu of a real way to name a function/function handle: 0;'])
F = @(x)HDC525(x);
and:
function HDC525(x)
%SCd 5/25/2011: Lame attempt at HD challenger
if ~isscalar(x)||x<=0
return
else
Str = evalc('whos x');
fprintf('%s\n ---- 0;'']) ------------------------------------------\n %s',Str(1:numel(Str)/2-1),Str(numel(Str)/2+1:end));
end

2 个评论

See Hint #1. Though I think you know this. I will post Hint #2 in a little while. This is a tough one!
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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