Passing Matrix using call by reference

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When looking for call-by-reference in matlab I found the Handle-Class. What I don't yet understand is how I use that class on a matrix so that I can pass a reference of the matrix and not the matrix itself.
I know matlab already kind of does this with large matrices, but firstly, I don't know how big my matrices will become, and secondly, I want to make sure it is only with reference, since it is done very often and is a performance critical operation.
Thank you.

Risposta accettata

Vaclav Rimal
Vaclav Rimal il 10 Dic 2013
To really take the advantage of handle class, you would need to create your own class based on the class handle. Create a file called largematrix.m having the following code:
classdef largematrix < handle
properties
array
end
methods
function myfunc(obj)
obj.array=obj.array+1;
end
end
end
Then, you can create an object of that class. The variable representing the object in the workspace is in fact a pointer to the part of memory which stores the object. Running the method myfunc behaves like you probably want. The following commands
a=largematrix;
a.array=ones(4);
a.myfunc; % or myfunc(a), which has the same result
a.array
produce
ans =
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2
  6 Commenti
Daniel
Daniel il 10 Dic 2013
True. Thanks for the tip.
Vaclav Rimal
Vaclav Rimal il 11 Dic 2013
Daniel, the methods will not be copied, for they are the same for all objects you create (assuming they are the same class). The methods should behave just like ordinary functions. I think this is not what matters. Matlab certainly deals this efficiently.
An issue can arise when you create and delete many instances. This depends on what exactly you want to do and whether you can aviod recreating objects by e.g. overwriting their properties by new values (which wouldn't repeatedly capture and release memory).

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Più risposte (4)

Jos (10584)
Jos (10584) il 10 Dic 2013
Yes, using assignin and input name will do that for you
function myfun(varargin)
assignin('caller',inputname(1),2)
>> test = 'hallo'
test =
hallo
>> myfun(test)
>> test
test =
2
  1 Commento
Daniel
Daniel il 10 Dic 2013
Thanks for the idea, I'll definitely follow up that path. It's just not what I was looking for.

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Sean de Wolski
Sean de Wolski il 10 Dic 2013
Modificato: Sean de Wolski il 10 Dic 2013
What operation are you doing inside of myfunc? Is it elementwise?
If so, MATLAB will do the operation in place if x is named the same everywhere:
x = magic(10);
x = myfunc(x);
%%
function x = myfunc(x)
x = x.^2;
end
Since the operation is being done in-place, no memory copy will be necessary.
  1 Commento
Daniel
Daniel il 10 Dic 2013
Operation isn't elementwise, but thanks for the info.

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Jos (10584)
Jos (10584) il 10 Dic 2013
Matlab does not create a copy, until it is really needed. So passing big arrays to functions uses no extra memory, except when you change the big matrix inside the function.
function myfun1(X)
Y = X ; No copy is made, Y is a pointer (a reference to X)
X(1) = 1 % this creates a copy of X inside the workspace of the function. Note that Y is untouched.
Y(1) = 1 ; % this now also creates a copy
  2 Commenti
Daniel
Daniel il 10 Dic 2013
Modificato: Sean de Wolski il 10 Dic 2013
I know this already. What I am looking for is a true call-by-reference ins this form:
x=1;
myFunc(x);
function myFunc(x)
x=2;
end
x
ans=2;
Is that possible?
Jos (10584)
Jos (10584) il 10 Dic 2013
Modificato: Jos (10584) il 10 Dic 2013
OK. Yes, it is possible. See my second answer.

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sixwwwwww
sixwwwwww il 10 Dic 2013
Define you function as below:
function y = myFunc(x)
y = 2;
end
then try this in command window:
x = 1;
x = myFunc(x);
It will return you value 2 which you need
  1 Commento
Daniel
Daniel il 10 Dic 2013
Sorry, but please read the question. I don't want call by value, which is what Matlab usually does in your case. Exception to this is in the case of large matrices, where there is a mix of both. Again not what I need and not reliable.
I did research the issue before asking: http://www.mathworks.com/matlabcentral/answers/96960
Please look into the issue to see if the obvious is what was asked for before answering.

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