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?
採用された回答
Laura Proctor
2011 年 5 月 25 日
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 件のコメント
その他の回答 (1 件)
Sean de Wolski
2011 年 5 月 25 日
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 件のコメント
Sean de Wolski
2011 年 5 月 25 日
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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